$p$-completion is pro-$p$ free Let $G$ be an abstract finitely generated residually finite group, and suppose that it's $p$-completion $\widehat{G_p}$ is a pro-$p$ free group. Does this implies that $G$ is a free group?

The converse is indeed true. For this proposition though, I'm unsure how to proceed. I'm using this result to prove that $H^2(G, \mathbb{F}_p) \neq 0 \implies H^2(\widehat{G_p},\mathbb{F}_p) \neq 0$. There seems to be no need for a basis of $\widehat{G_p}$ to be contained in $G$, or for one such basis to exist. A density argument may do something here, but I'm a bit out of ideas.

I believe this to be true given that pro-$p$ completion commutes with group presentations in a sense - the abstract presentation of $G$ is a topological presentation of $\widehat{G_p}$. The converse of such statement would be a proof of the result I'm looking for.
This is false. All parafree groups have this property. Look at https://en.wikipedia.org/wiki/Parafree_group

One of the easiest example of a parafree group is 

$\langle x,y,z: x^2y^2=z^3\rangle$.

This group is also residually-$p$ for all primes $p$.
This is false. All parafree groups have this property. Look at https://en.wikipedia.org/wiki/Parafree_group

One of the easiest example of a parafree group is 

$\langle x,y,z: x^2y^2=z^3\rangle$.

This group is also residually-$p$ for all primes $p$.