And it would also be true that $\tilde{F}$ is surjective iff $F$ is surjective on objects, but that didn't fit in the title.
For $F: \mathcal{C} \to \mathcal{D}$, we have a functor $\widehat{\mathcal{C}} \to \widehat{\mathcal{D}}$ given by $FX(D) = \widehat{\mathcal{C}}(\hom(F(-), D), X)$. This is right adjoint to $- \circ F$.
I feel we should either look at its adjoint or use something similar to the Yoneda lemma. For the right adjoint, we have that $- \circ F$ is surjective if $F$ is surjective on objects because for $Y: \mathcal{C} \to Set$ we can define $XD$ to be equal to $YFC$ for some $C$ such that $FD = C$ and then $YF = X$. I don't think it is true that a left adjoint being surjective implies that the left one is surjective because I have not been able to find that on google and it seems like a fact that would not be obscure if it's true.
For simplicity, I'll assume $\mathcal C$ and $\mathcal D$ are small. Then a functor $F \colon \mathcal C \to \mathcal D$ has a nerve $N_F \colon \mathcal D \to \widehat{\mathcal C}$, which is dense if and only if $F$ is fully faithful (Remark 7.2 of Avery–Leinster's Isbell conjugacy and the reflexive completion). Furthermore, the nerve $N_F$ itself has a nerve $N_{N_F} \colon \widehat{\mathcal C} \to \widehat{\mathcal D}$, which is fully faithful if and only if $N_F$ is dense (by definition of density). $N_{N_F}$ is precisely the functor $\tilde F$ you describe. Thus $\tilde F$ is fully faithful if and only if $F$ is fully faithful.