No, being torsion is not a local property, and I can give a counterexample. [Edit: This took some doing, with my initial answer containing a serious flaw. After completely reworking the construction, this should work now. Apologies for the length of this answer, but I don't see any quick constructions].
The idea is to construct a ring $R$ and an ideal $I$ contained in the zero divisors of $R$, and $f_1,f_2in R$ satisfying $f_1+f_2=1$ such that $I_{f_i}$ contains a regular element of $R_{f_i}$ for each $i$.
This provides a counterexample to the question by taking the module $M=R/I$ and $m=I+1$. Then, ${rm ann}(m)=I$ consists of zero divisors, so $m$ is not torsion. However, mapping $m$ into $M_{f_i}$ takes ${rm ann}(m)$ to $I_{f_i}$, which contains regular elements of $R_{f_i}$. So, $m$ is torsion in each $M_{f_i}$.
This does get rather involved, so let's start simple and construct an example showing that being torsion is not a stalk-local property.
Choose a field $k$, set $A=k[X_0,X_1,X_1,X_2,ldots]$ and let $J$ be the ideal generated by $X_iX_j$ for $inot=j$ and $X_i(X_i-1)$ for $ige1$. Then, $R=A/J$ is the $k$-algebra generated by elements $x_0,x_1,ldots$ satisfying the relations $x_ix_j=0$ for $inot=j$ and $x_i(x_i-1)=0$ for $ige1$. Let $Isubseteq R$ be the ideal generated by $x_0,x_1,ldots$.
We can see that $x_inot=0$ by considering the $k$-morphism $Ato k$ taking $X_j$ to 1 (some fixed $j$) and $X_i$ to 0 for $inot=j$. This takes $J$ to 0, so it defines a $k$-morphism $Rto k$ mapping $x_j$ to 1, so $x_jnot=0$. Then, every $ain I$ satisfies $ax_j=0$ for large $j$, showing that it is a zero divisor. Also, the $k$-morphism $Ato k$ taking each $X_i$ to zero contains $J$ in its kernel, and defines a morphism $Rtomathcal{k}$ with kernel $I$, showing that $R/Icong k$. So $I$ is a maximal ideal. For any prime $mathfrak{p}$ we either have $mathfrak{p}not=I$, in which case the non-empty set $Isetminusmathfrak{p}$ maps to units (and hence, regular elements) in $R_{mathfrak{p}}$. Or, we have $mathfrak{p}=I$ in which case $x_i-1$ maps to a unit and $x_i$ goes to zero in $R_{mathfrak{p}}$ ($ige1$). So, $R_{mathfrak{p}}cong k[X]$ with $x_0$ going to the regular element $X$. This shows that $I$ contains regular elements in the localization at any prime, giving the required counterexample for the stalk-local case.
Now, let's move on to the full construction of the counterexample showing that being torsion is not a local property. Simply guessing a set of generators and relations as for the stalk-local case didn't work out so well. Instead, I will start with a simple example of a polynomial ring and then transform it in such a way as to give the properties we are looking for. I find it helpful to first fix the following notation: Start with the base (polynomial) ring $R=mathbb{Z}[x,y,z]$. A (commutative, unitial) R-algebra is simply a ring with three distinguished elements $x,y,z$, and a morphism of R-algebras is just a ring homomorphism respecting these distinguished elements. For an R-algebra $A$, define $K(A)subseteq A$ to be the smallest ideal such that, for all $ain A$,
$$
begin{align}
axin K(A)&Rightarrow azin K(A),\\
ayin K(A)&Rightarrow a(1-z)in K(A).
end{align}
$$
In particular, $K(A)=0$ implies that $x$ is a regular element in the localization $A_z$ and $y$ is a regular element in $A_{1-z}$. If we can construct such an example where the ideal $Ax+Ay$ consists purely of zero divisors, then that will give the counterexample needed. The idea is to start with $A=mathbb{Z}[x,y,z]$ and transform it using the following steps.
- Force the elements of $I=Ax+Ay$ to be zero divisors. So, for each $ain I$, add an element $b$ to $A$ in as free a way as possible such that $ab=0$. Adding elements to $A$ also has the effect of adding elements to $I$. So, this step needs to be iterated to force these new elements of $I$ to also be zero divisors.
- Replace $A$ by the quotient $A/K(A)$ to force the condition $K(A)=0$.
The first step above is easy enough. However, we do need to be careful to check that the second step does not undo the first. Suppose that $ain A$ is a zero divisor, so that $ab=0$ for some non-zero $b$. It is possible that taking the quotient in the second step above takes $b$ to zero, so that $a$ becomes a regular element again. To get around this, we need some stronger condition on $b$ which implies $bnot=0$ and is also stable under taking the quotient. Note that $A(1-b)$ being a proper ideal or, equivalently, $A/(1-b)$ being nontrivial, will imply that $bnot=0$. In turn, this is implied by $K(A/(1-b))$ being a proper ideal. As it turns out, this property of $b$ does remain stable under each of the steps above, and can be used to show that this construction does give the counterexample required. However, note that if $ab=0$ and $K(A/(1-b))$ is proper, then $a=a(1-b)in A(1-b)$, from which we can deduce that $K(A/(a))$ is a proper ideal. This necessary condition is unchanged by either of the steps above, so we had better check that elements $ain Ax+Ay$ in our R-algebra do satisfy this from the outset. I'll make the following definition: $A$ satisfies property (P) if $K(A/(a))$ is proper for every $ain Ax+Ay$. As it turns out, polynomial rings do satisfy this property and, consequently, the construction outlined above works fine.
Now on to the details of the argument.
(1) Let $fcolon Ato B$ be an R-morphism. Then $f(K(A))subseteq K(B)$. Furthermore,
- If $Isubseteq A$, $Jsubseteq B$ are ideals with $f(I)subseteq J$ and $K(B/J)$ is proper, then $K(A/I)$ is proper.
- If $B$ satisfies (P) then so does $A$.
As $f^{-1}(K(B))$ satisfies the defining properties for $K(A)$ (other than minimality), it contains $K(A)$. In particular, if $K(B)$ is proper then $K(A)subseteq f^{-1}(K(B))$ is proper. Next, if $f(I)subseteq J$ are ideals, then $f$ induces an R-morphism $A/Ito B/J$ so, if $K(B/J)$ is proper then so is $K(A/I)$.
If $B$ satisfies property (P) and $ain Ax+Ay$ then $f(a)in Bx+By$ and $K(B/f(a))$ is proper. So, $K(A/(a))$ is proper and $A$ also satisfies property (P).
(2) If $A$ is a non-trivial ring then the polynomial ring $Rotimes Acong A[x,y,z]$ satisfies (P).
As $A$ is non-trivial, it must have a maximal ideal $mathfrak{m}$. Applying (1) to the R-morphism $A[x,y,z]to(A/mathfrak{m})[x,y,z]$ reduces to the case where $A=k$ is a field. Then, letting $bar k$ be the algebraic closure, applying (1) to $k[x,y,z]tobar k[x,y,z]$ reduces to the case where $A=k$ is an algebraically closed field.
Now set $B=k[x,y,z]$ and choose $ain Bx+By$. The idea is to look at the morphism $thetacolon B/(a)to k$ taking $x,y,z$ to some $x_0,y_0,z_0in k$ with $a(x_0,y_0,z_0)=0$. As long as these satisfy $ux_0=0Rightarrow uz_0=0$ and $uy_0=0Rightarrow u(1-z_0)=0$ (all $uin k$) then $K(B/(a))$ will be contained in the kernel of $theta$, so will be proper. For this to be the case it is enough that both ($x_0not=0$ or $z_0=0$) and ($y_0not=0$ or $z_0=1$).
Case 1: We can find $a(x_0,y_0,z_0)=0$ such that $x_0y_0not=0$. This satisfies the required condition.
Case 2: Whenever $a(x_0,y_0,z_0)=0$ then $x_0y_0=0$. This means that $xy$ is contained in the radical ideal generated by $a$, so $a$ divides $x^ry^r$ some $rge1$. Then $a$ is a multiple of $x$ or $y$ and one of $(x_0,y_0,z_0)=(0,1,0)$ or $(1,0,1)$ satisfies the required condition. So, $K(B/(a))$ is proper.
Next, we construct extensions of the R-algebra forcing elements of $Ax+Ay$ to be zero-divisors.
(3) If $A$ satisfies (P), then we can construct an R-morphism $fcolon Ato B$ with a left-inverse and such that, for every $ain Ax+Ay$, there is a $bin B$ with $ab=0$ and $K(B/(1-b))$ is proper.
To construct the morphism, set $I=Ax+Ay$ and let $(X_a)_{ain I}$ be indeterminates over $A$. Let $J$ be the ideal in $A[(X_a)_{ain I}]$ generated by $(aX_a)_{ain I}$. Then define $B=R[(X_a)_{ain I}]/J$ and let $f$ be the canonical homomorphism. Its left inverse is the map taking $X_a$ to 0.
Now, for a fixed $ain I$, set $b=J+X_a$, so $ab=0$. Consider the morphism $A[(X_c)_{cin I}]to Ato A/(a)$ taking each $X_c$ to 0 (for $cnot=a$) and $X_a$ to 1. As its kernel contains $J$, it defines a morphism $gcolon Bto A/(a)$, which takes $b$ to one. Therefore, the ideal $B(1-b)$ maps to 0 and, as $K(A/(a))$ is proper, (1) says that $K(B/(1-b))$ is proper.
(4) If $A$ satisfies (P), then we can construct an R-morphism $fcolon Ato B$ such that, for every $ain Bx+By$ there is a $bin B$ with $ab=0$ and $K(B/(1-b))$ is proper.
Set $A_0=A$ and use (3) to construct a sequence of extensions $f_icolon A_ito A_{i+1}$ with left inverses such that, for every $ain A_ix+A_iy$ there is a $bin A_{i+1}$ with $ab=0$ and $K(A_{i+1}/(1-b))$ is proper. Note that, as each $f_i$ has a left inverse, (1) says that $A_{i+1}$ satisfies (P) whenever $A_i$ does. So, we can keep applying (3) to build up the entire sequence of extensions.
We now take the colimit $B={rm colim}A_i$ and let $f$ be the induced morphism (i.e, if we consider $A_isubseteq A_{i+1}$ then $B$ is the union and $f$ is inclusion). As each $A_ito B$ has a left-inverse, (1) shows that the required properties for $B$ are inherited from the individual $A_i$.
(5) Suppose that $A$ satisfies the following: for every $ain Ax+Ay$ there is a $bin A$ with $ab=0$ and $K(A/(1-b))$ is proper. Then, the R-algebra $B=A/K(A)$ satisfies the same property, and also $K(B)=0$.
That $K(B)=0$ follows quickly from the definition of $K$.
Suppose $ain Ax+Ay,bin A$ are such that $ab=0$ and $K(A/(1-b))$ is proper. Set $C=A/(1-b)$, so that $C/K(C)$ is not trivial. By (1), the canonical morphism $Ato C$ maps $K(A)$ into $K(C)$. So, it induces a morphism $Bto C/K(C)$. This takes $1-b$ to zero, so it induces an R-morphism $B/(1-b)to C/K(C)$. As $K(C/K(C))=0$, (1) implies that $K(B/(1-b))$ maps to zero, so is proper.
(6) If $B$ is the R-algebra constructed in (5), then $I=Bx+By$ contains only zero-divisors but $x,yin I$ map to regular elements in $R_z$ and $R_{1-z}$ respectively.
For any $ain I$ there is a $bin B$ with $ab=0$ and $K(B/(1-b))$ proper. In particular, $(1-b)$ must be a proper ideal, so that $bnot=0$ and $a$ is a zero divisor.
Finally, the property $K(B)=0$ implies that $x$ is regular in $B_z$ and $y$ is regular in $B_{1-z}$.
