If the fourier transformed of f is odd, is f odd?

Look at $L^2$ first: In $L^2$ the FT is diagonalizable. The space of odd functions $in L^2$ is the direct sum $Eig(mathcal{F},+i)oplus Eig(mathcal{F},-i)$ of the eigenspaces of $mathcal{F}$ with respect to the eigenvalues $+i$ and $-i$. Because eigenspace are mapped into themselves, $mathcal{F}$ maps odd functions to odd functions.

In particular this is true for all $fin L^1cap L^2$ and by continuity it is true for all $fin L^1$. (In fact an analogue statement is true for all tempered distributions.)

EDIT: Oh, I just saw that you asked for the other direction. Using the same argument you can show that $mathcal{F}f$ odd $implies f=mathcal{F}^3(mathcal{F}f)$ odd is true for all $fin L^2$. This time I'm not quite sure if it is possible to extend this from $L^1cap L^2$ to $L^1$, but maybe the result for $L^1cap L^2$ is useful for you too.