Suppose that $x$ is an object of $\mathcal{G}$ over a finite extension $k'/k$. Denote $\sigma : k' \to k'$ the $q$-power frobenius. Let $x^\sigma$ be the pullback of of $x$ by $\sigma$. As $\mathcal{G}$ is a gerbe over $k$, after replacing $k'$ by a further finite extension, we may assume there is an isomorphism $\alpha : x^\sigma \to x$ over $k'$. If $n = [k' : k]$, then consider the automorphism $\beta = \alpha \circ \alpha^\sigma \circ \dotsb \circ \alpha^{\sigma^{n - 1}}$ of $x$ over $k'$. If $\beta = \operatorname{id}$, then $\alpha$ determines a descent datum and since $\mathcal{G}$ is a stack, we would be able to descend $x$ to an object over $k$ and the gerbe would be neutral.

OK, but perhaps $\beta$ is not trivial. However, then $\beta$ is an element of the automorphism group of $x$ over $k'$ which is (if we have a suitable finiteness assumption on $\mathcal{G}$, for example if $\mathcal{G}$ is quasi-separated — an assumption that always holds in practice) the $k'$-points of an algebraic group over $k'$. Thus $\beta$ has finite order as $k'$ is a finite field. Say $\beta$ has order $m \geq 1$. Then after replacing $k'$ by an extension of degree $m$, and going trough the whole process again, we end up with $\beta = \operatorname{id}$ and $\mathcal{G}$ is neutral.