It is not true : for instance, $X$ and $X'$ might be two non-isomorphic $G$-torsors on the same $S$-scheme. S$-scheme $T$.
Here is a precise example. Consider $S=Spec(\mathbb{R})$ S=T=Spec(\mathbb{R})$ and $G=\mathbb{Z}/2\mathbb{Z}$, let $X=Spec(\mathbb{C})$ and $X'=Spec(\mathbb{R})\cup Spec(\mathbb{R})$ viewed as $S$-schemes, and let $G$ act on $X$ and $X'$ respectively by complex conjugation and by exchanging the connected components. Both actions admit $Spec(\mathbb{R})$ as a geometric quotient. But an isomorphism between these quotients obviously doesn't lift to an equivariant morphism between $X$ and $X'$.
Here is another example : take $S=Spec(\mathbb{C})$, $G=\mathbb{G}_m$, $T=\mathbb{P}^1_{\mathbb{C}}$. Let $X$ (resp. $X'$) be the total space of the line bundle $\mathcal{O}$ (resp. $\mathcal{O}(1)$) on $T$ minus the zero-section, with the natural $G$-action. Both actions admit $T$ as a geometric quotient, but there is no equivariant morphism between $X$ and $X'$ lifting identity.
