Perhaps this is not the strongest result one can get, but it is true that, if $M$ is a complete Riemannian manifold, then for almost all $p^*\in M$ the set $F_{p^*}$ you define in the question has measure $0$. Due to Fubini's theorem, we have:

For almost all $p^*\in M$ the set $F_{p^*}$ has measure $0$.

iff

$$\mu\times\mu(\{(p,p^*)\in M\times M;\text{there is a unique minimizer of $d(p^*,\cdot)$ in }O_p\})=0.$$

iff

For almost all $p\in M$, the set $\{p^*\in M;\text{there is a unique minimizer of $d(p^*,\cdot)$ in }O_p\}$ has measure $0$.

But the last statement is in fact satisfied for all $p\in M$; indeed, $O_p$ is closed for all $p$, because the action of $G$ on $M$ is proper, so by [the answer](https://mathoverflow.net/a/469177/172802) I gave to your other question, we are done.