Characterizing the Radon transforms of log-concave functions $f:\mathbf{R}^d\to \mathbf{R}_{\ge 0}$ is log-concave if $\log(f)$ is concave (and the domain of $\log(f)$ is convex).
Theorem: For all $\sigma$ on the sphere $\Bbb S^{d-1}$ and $r\in \mathbf{R}$,
$$
g_\sigma(r) := \int\limits_{\sigma\cdot x=r}f(x)\,\mathrm{d}S(x)
$$
is a log-concave function of $r$. (Note: $g$, as a function of $\sigma$ and $r$, is the Radon transform of $f$.)
Question: does this characterize log-concavity? That is, if $g_\sigma(r)$ is log-concave as a function of $r$ for all $\sigma$, is $f$ log-concave?
 A: The following answer exploits a wider context, where non-negative functions $f$, viewed as densities of absolutely continuous measures $f(x)dx$, are replaced by non-negative measures $\mu$. Because a concave function unbounded from above is $\equiv+\infty$, a log-concave measure $\mu$ is naturally an absolutely continuous measure with log-concave density. 
Therefore the Lebesgue measure over the $2$-dimensional sphere $S^2$ is not log-concave over ${\mathbb R}^3$. Nevertheless, its Radon transform in a direction $\sigma$ is
$$g_\sigma=2\pi\chi_{(-1,1)},$$
which is log-concave for every $\sigma$.
A: If I understand the question correctly, I think the answer is no.
Start with the following : if $f$ is the indicator function of the unit ball, then the function $g_\sigma(r)$ is strictly log-concave close to 0 (this function does not depend on $\theta$).
Now, let $h$ be the indicator function of the ball of radius $r<1$. Then $f-\epsilon h$ is never log-concave for any $\epsilon>0$, and its Radon transform (which again is independent of $\theta$) remains log-concave if epsilon is small enough.
(this is especially easy to see in dimension 2, in which case the Radon transforms of both $f$ and $h$ are second-degree polynomials on their support)
