Yes, this is true.
First, note that any concave PDF must be unimodal. (This is an elementary consequence of the concavity.)
The answer to a related question shows that on the interval $[0,1]$, the variance of a distribution with mean $\mu$ cannot exceed $\mu(2-3\mu)/3$ (when $0 \le \mu \le 1/2$) or $(1-\mu)(3\mu-1)/3$ (when $1/2\le \mu \le 1$). Neither of these expressions exceeds $1/12$ (which is achieved solely when $\mu=1/2$.
By shifting and rescaling the interval $[0,1]$ to $[a,b]$, it immediately follows that $\sigma^2$ cannot exceed $(b-a)^2/12$, QED.
Evidently a stronger statement can be made when the mean $\mu$ is known: a tight upper bound is then
$$\sigma^2 \le \frac{(\mu-a)(2b+a-3\mu)}{3}$$
when $\mu \le (a+b)/2$. For $\mu \gt (a+b)/2$, switch $a$ and $b$ and replace $\mu$ by $b-\mu$.