homological algebra - Is Tor always torsion?

Yes. Set $K = mathrm{Frac} R$.

Lemma: Let $ldots to C_2 to C_1 to C_0$ be a complex of $R$-modules. Suppose that $C^{bullet} otimes_R K$ is exact (but not necessarily surjective at $C_0$). Then $H_k(C_{bullet})$ is $R$-torsion for $k>0$.

Proof: Let $v in C_k$ with $dv=0$. So $d(v otimes 1)=0$. By the exactness of $C^{bullet} otimes_R K$, there is $u in C_{k+1} otimes_R K$ with $du=v$. Write $u=sum w_i otimes (f_i/g_i)$, with $f_i/g_i in K$ and $w_i in C_{k+1}$. Set $g=prod g_i$ and $w=sum left( prod_{j neq i} g_j right) f_i w_i$. Then $dw=gv$, so $[v]$ is $g$-torsion in $H_k(C_{bullet})$. QED

Take resolutions $A_{bullet} to M$ and $B_{bullet} to N$ by free $R$-modules. Then $mathrm{Tor}_{bullet}(M,N)$ is the homology of the complex formed by collappsing the double complex $A_{bullet} otimes_R B_{bullet}$. Note that $left( A_{bullet} otimes_R B_{bullet} right) otimes_R K cong (A_{bullet} otimes_R K) otimes_K (B_{bullet} otimes_R K)$.

Since $A^{bullet}$ is exact, so is $A^{bullet} otimes_R K$. Thus $A_{bullet} otimes_R K$ breaks up as a direct sum of complexes of the form $ldots ldots 0 to K to K to 0 ldots$, and the complex $ldots to 0 to K$, with the $K$ in the last position. (This uses the Axiom of Choice; I suspect you should be able to avoid it, but I don't see how right now.) The complex $B otimes_R K$ breaks up into pieces of the same kind. So the double complex breaks up into squares $begin{smallmatrix} K & to & K \ downarrow & & downarrow \ K & to & K end{smallmatrix}$, vertical strips $begin{smallmatrix} K \ downarrow \ K end{smallmatrix}$, horizontal strips $begin{smallmatrix} K & to & K end{smallmatrix}$ and, in position $(0,0)$, some isolated copies of $K$.

Only summands of the last type contribute to the cohomology of the double complex, so the double complex obeys the hypotheses of the lemma and we are done.