This question could be easily disproven by a counterexample or easily proven, but i cant think of any...
In the filtered probability space $(\Omega,\mathcal{A},\{\mathcal{F}_t:t\in\mathbb{R}\},\mathbb{P})$ choose a $C\in \mathcal{F}_\infty = \sigma\bigcup_t \mathcal{F}_t$ and a sub-$\sigma$-field $\mathcal{A}_0\subset\mathcal{A}$ independant to $\mathcal{F}_\infty$. Set $\mathcal{G}_t = \sigma(\mathcal{F}_t \cup \mathcal{A}_0)$, i.e. $\{\mathcal{G}_t : t\in\mathbb{R}\}$ is an enlarged filtration and every $\mathcal{F}$-stopping time $\tau$ is also a $\mathcal{G}$-stopping time. For $t\in\mathbb{R}$, $F\in\mathcal{F}_t$ and $A\in\mathcal{A}_0$ one gets (since independant r.v. are uncorrelated)
\begin{equation} \mathbb{E}\left[ \mathbb{P}\left[ \left. C \right\vert \mathcal{F}_t \right]; F\cap A \right] = \mathbb{E}\left[ \mathbb{P}\left[ \left. C \right\vert \mathcal{F}_t \right]; F \right] \mathbb{P}[A] = \mathbb{P}[C\cap F\cap A] \end{equation} and thus $ \mathbb{P}[C\vert \mathcal{F}_t] \stackrel{\text{as}}{=} \mathbb{P}[C\vert\mathcal{G}_t] $ due to a monotone class argument (could be a standard result), since $ \{ \Omega_0\in\mathcal{A} : \mathbb{E}\left[ \mathbb{P}\left[ \left. C \right\vert \mathcal{F}_t \right]; \Omega_0 \right] = \mathbb{P}[C\cap\Omega_0] \} $ contains $\Omega$ and is closed under relative complements and countable disjoint unions, and all the sets $F\cap A$ form a generator of $\mathcal{G}_t$, which is closed under pairwise intersection.
Can one also deduce $\mathbb{P}[C\vert\mathcal{F}_\tau] \stackrel{\text{as}}{=} \mathbb{P}[C\vert\mathcal{G}_\tau]$ ? Is the above argument (for $\tau\equiv t$) right, or did i make an error ? Here $\mathcal{F}_\tau$, resp. $\mathcal{G}_\tau$ are the stopped (wrt $\tau$) $\sigma$-fields, i.e. all $A\in\mathcal{A}$ with $A\cap\{\tau\leq t\}\in\mathcal{F}_t$, resp. $A\cap\{\tau\leq t\}\in\mathcal{G}_t$ for all $t\in\mathbb{R}$.
Help is appreciated and gladly accepted.