If an automorphism fails (C), then obviously it satisfies (B). An (A)+(B) example easily extends from rank 1 to ranks 2: take a root of unity $\zeta$ in $\mathbb Z_p^\times$ that is not $\pm1$ (requiring $p\geqslant 5$). If the generators of the free group are $x$ and $y$, then there is an automorphism of the $p$-adic completion given by $x\mapsto \zeta x$ and $y\mapsto y$. It obviously has finite order, but it has nontrivial determinant, so it is not in the closure of $Out(F)$.

Define $SOut$ as the subgroup of $Out$ with determinant $1$. I believe that $SOut(F)$ is dense in $SOut(\hat F)$, so there is nothing satisfying (B)+(C), let alone all three conditions. I will not attempt to show that, but only that all torsion is conjugate into the closure, which is awfully close. Specifically, I will show that an $\ell$-Sylow subgroup is contained in the closure; and I will suggest that there is no $p$-torsion ($p\geqslant 5$). (Added at last minute: that only covers prime-power torsion and does not show that torsion of order with multiple prime factors lifts from $SL_2(\mathbb Z_p)$ to $SOut(\hat F)$, let alone to the closure of $SOut(F)$.) The key point is that $\hat F$ is a pro-p-group and thus pro-nilpotent. That makes it much easier to analyze than if it were, say, the 2,3-completion and pro-soluble, let alone the full completion with simple composition factors. The length $n$ nilpotent quotient of a group is a characteristic quotient, so an automorphism of a group induces a automorphism of its length $n$ nilpotent quotient. The group of automorphisms of a pro-nilpotent group is the inverse limit of the automorphism groups of the length $n$ quotients. (The transition maps in this inverse limit need not be surjective, but are in the free case, as indicated below.)

The automorphism group of a nilpotent group is easy to understand. If $G_n$ is a length $n$ nilpotent group, $G_{n-1}$ its maximal length $n-1$ quotient and $Z$ the kernel, then the kernel of $Aut(G_n)\to Aut(G_{n-1})$ consists of automorphisms that differ from the identity by shearing into the kernel: $\phi$ so that $\phi(g)g^{-1}\in Z$. Using the centrality of $Z$, the map $g\mapsto \phi(g)g^{-1}$ is a homomorphism $G_n\to Z$. A useful observation is $\phi(g)g^{-1}=g^{-1}\phi(g)$. The kernel is isomorphic to to the group of homomorphisms $G_n^{ab}\to Z$. $Aut(G_n)$ need not surject to $Aut(G_{n-1})$, but a similar analysis shows that if an automorphism of $G_{n-1}$ lifts to an endomorphism of $G_n$, it has an inverse. By induction that yields a nice clean statement: an endomorphism of a nilpotent group is an automorphism if and only if the induced endomorphism of the abelianization is an automorphism.

Applied to $\hat F$, this shows that the kernel $Aut(\hat F)\to GL_2(\mathbb Z_p)$ is a torsion-free pro-$p$ group. Also, the map is surjective because freeness makes it easy to lift endomorphisms, which are then automorphisms. (Similarly, it is easy to lift automorphisms of the length $n$ nilpotent quotient to endomorphisms of the free group, but it takes work to lift them to automorphisms, which is why I do not do it. That would imply that $SOut(F)$ is dense in $SOut(\hat F)$. That would answer your question, but the points about Sylow subgroups would still be interesting.)

Thus the kernel $Aut(\hat F)\to GL_2(\mathbb Z_p)$ is built out of composition factors of the form $Hom(A,B)$, where $A$ and $B$ are composition factors of $\hat F$, so the kernel is a pro-$p$ group. More careful consideration shows that $A$ and $B$ are torsion free, hence so the kernel. The kernel $Out(\hat F)\to GL_2(\mathbb Z_p)$ is a quotient of the prior kernel by $\hat F$, so also a pro-$p$ group. I believe that directly applying an analogous stage-by-stage analysis shows that it is torsion-free. The kernel of $GL_2(\mathbb Z_p)\to GL_2(\mathbb F_p)$ is also a pro-$p$-group, torsion-free unless $p=2$. Extension by a $p$-group cannot change the prime-to-$p$ Sylow subgroups. For each $\ell$ prime to $p$, the $\ell$-Sylow subgroups of $Aut(\hat F)$, $Out(\hat F)$, $GL_2(\mathbb Z_p)$, and $GL_2(\mathbb F_p)$ are isomorphic by the quotient map. Similarly, the $\ell$-Sylow subgroups of $SAut(\hat F)$, $SOut(\hat F)$, $SL_2(\mathbb Z_p)$, and $SL_2(\mathbb F_p)$ are isomorphic. Since $SOut(F)=SL_2(\mathbb Z)$ surjects to $SL_2(\mathbb F_p)$, its closures in $SAut(\hat F)$, $SOut(\hat F)$, and $SL_2(\mathbb Z_p)$ must contain full $\ell$-Sylow subgroups. In particular, all $\ell$-power torsion is conjugate into Sylow subgroups and thus into the closure of $SOut(F)$. And I claim that there is no $p$-power torsion (except for $p=2,3$, where all the torsion in $GL_2(\mathbb Z_p)$ is conjugate into $GL_2(\mathbb Z)=Out(F)$).

All that applies to any rank free group ($Out(F)$ always surjects to $GL(\mathbb Z)$, but is no longer a isomorphic), except that there are more possibilities for $p$-power torsion in $GL_n(\mathbb Z_p)$. I think it is all conjugate into $GL_n(\mathbb Z)$ (though I’m more sure about the fields $\mathbb Q_p$ and $\mathbb Q$). I don’t know how to tell if it lifts to $Out(\hat F)$, but if it does, it probably lifts to $Out(F)$. Anyhow, I claim without proof that $SOut(F)$ is dense in $SOut(\hat F)$.