There does, but it's not such an obvious fact.

 Let $\tau$ be an $(\mathcal F_t)$ stopping time. Then $\tau$ is also a stopping time of the augmented filtration $(\tilde {\mathcal F}_t)$, obtained by adding all null sets in the completion of $\mathcal F_\infty$ to each $\mathcal F_t$. As noted already, it is a consequence of the Blumenthal zero-one law that $(\tilde{ \mathcal F}_t)$ coincides with the analogous augmentation $(\tilde{\mathcal G}_t)$ of $(\mathcal G_t)$. Also, it is known (see section 5 of Chapter IV of the book of Revuz and Yor, for example) that each $(\tilde{ \mathcal F}_t)$ stopping time is predictable. In particular, $\tau$ is $(\tilde{ \mathcal F}_t)$-predictable. Thus the process $Z_t:=1_{\{\tau=t\}}$ is $(\tilde{ \mathcal G}_t)$-predictable. By the "de-completion" result stated as a Lemma in section 6 of Appendix I of Volume B of *Probabilités et Potential* by Dellacherie and Meyer, $Z$ is indistinguishable from a $(\mathcal G_t)$-predictable process $Z'$. The $(\mathcal G_t)$ stopping time $\tau':=\inf\{t:Z'_t=1\}$ is a.s. equal to $\tau$.