I don't know a reference, but here's the rough proof that $G \in C^1$.
First, from the integral eqaulity $$ \Phi_t(x+h) - \Phi_t(x) = \int_0^1 D\Phi_t(x+\eta h) \cdot h \,\mathrm{d}\eta $$ follows the mean-value estimate $$ \| \Phi_t(x+h) - \Phi_t(x) - D\Phi_t(x) \cdot h \| \le \sup_{\eta \in [0,1]} \| D\Phi_t(x+\eta h) - D\Phi_t(x) \|\,\|h\|. $$ This allows us to estimate $$ \|G(x+h) - G(x) - \int_0^\infty D\Phi_t(x) \cdot h \,\mathrm{d}t\| \\ \le \int_0^\infty \| \sup_{\eta\in[0,1]} D\Phi_t(x+\eta h) - D\Phi_t(x) \|\,\mathrm{d}t\,\|h\|. $$ To show that this is $o(h)$, we split up the integral into intervals $[0,T]$ and $[T,\infty)$. For any finite $T$, the first part is small since $t \mapsto D\Phi_t(x)$ is uniformly continuous on $[0,T]$. The second part is exponentially small when $T \to \infty$ due to the exponential stability. Hence $G$ is differentiable and $$ DG(x) = \int_0^\infty D\Phi_t(x) \,\mathrm{d}t. $$ Similar estimates can be repeated to prove that $G \in C^1$ and by induction that $G \in C^k$.