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 this space$Y$ 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 $\text{Hom}(G,\Gamma\rtimes G)$ consistingwhich is the preimage of homomorphisms with full image after projecting to $G$$Y$. It is enough to show We know, by local rigidity, that $X$$\text{Hom}(G,\Gamma\rtimes G)$ is discrete mod the action of $\text{Aut}(G)\times \Gamma$, where the first factor acts by pre-composition and the second$\Gamma\rtimes G$ by post-composition via itsthe inner action. In fact Thus also $X$, itwhich is already discrete moda subspace $G\times \Gamma$$\text{Hom}(G,\Gamma\rtimes G)$, where and $G$ acts by precomposition via its inner action. Indeed$Y$, by the defining propertywhich is a qoutient of $X$, are discrete mod the action of $G\times \Gamma$-orbits$\Gamma\rtimes G$. We are the same as thethus left to show that $\Gamma$ acts transitively on each $\Gamma\rtimes G$-orbits fororbit in $Y$. Note that every element in $y\in Y$ is a subgroup of $\Gamma\rtimes G$ satisfying $\Gamma\cdot y=\Gamma\rtimes G$ and the postcomposition innersubgroup $y$ clearly stabilizes the element $y$ by the conjugation action, so this follows from. Thus indeed $\Gamma$ acts transitively on the discretenessorbit of $\text{Hom}(G,\Gamma\rtimes G)$ mod $G\rtimes \Gamma$$y$.