Yes, this is true. 

The space of crossed homomorphisms from $G$ to $\Gamma$ could be identified, by taking graphs, with the space $Y$ consisting of subgroups of $\Gamma\rtimes G$ which intersect $\Gamma$ trivially and project onto $G$. You ask for discreteness of $Y$ modulo the $\Gamma$-conjugation action. 
Every such subgroup is the image of an homomoprhism $G\to \Gamma\rtimes G$, thus may be identified as an element of the space $\text{Hom}(G,\Gamma\rtimes G)$ modulo the action of $\text{Aut}(G)$ by pre-composition. Let me write $X$ for the subspace of $\text{Hom}(G,\Gamma\rtimes G)$ which is the preimage of $Y$. 
We know, by local rigidity, that $\text{Hom}(G,\Gamma\rtimes G)$ is discrete mod the action of $\Gamma\rtimes G$ by post-composition via the inner action. 
Thus also $X$, which is a subspace $\text{Hom}(G,\Gamma\rtimes G)$,
and $Y$, which is a qoutient of $X$, are discrete mod the action of $\Gamma\rtimes G$.
We are thus left to show that $\Gamma$ acts transitively on each $\Gamma\rtimes G$-orbit in $Y$. Note that every element in $y\in Y$ is a subgroup of $\Gamma\rtimes G$ satisfying $\Gamma\cdot y=\Gamma\rtimes G$
and the subgroup $y$ clearly stabilizes the element $y$ by the conjugation action. Thus indeed $\Gamma$ acts transitively on the orbit of $y$.