I am not convinced. Say that we consider continuity of $x↦E[H(x)|G]$ instead of differentiability, and $H(x)$ uniformly integrable. Then we know that if $x_n\rightarrow x$ then $E[H(x_n)|\mathcal{G}]\rightarrow E[H(x)|\mathcal{G}]$, P-a.s. Let $A_{x_n\rightarrow x}$ be the set of $\omega$ where this convergence does not hold. Now, there is an uncountable number of sequences converging to each $x$, and an uncountable number of $x$′s, and to have $x\mapsto E[H(x)|\mathcal G](\omega)$ contiuous for P−a.e.$\omega$, we also need that $P(\cup_x\cup_{x_n\rightarrow x} A_{\{x_n\},x} )=0$.