Yes. Here's a recipe to get such a group. Find a profinite group $K$ with three elements $a',b',c'$ such that $\langle a',b'\rangle$ is dense and $(a',b',c')$ is a free family. If so, we can "pull them back" to $\hat{F_2}$. Namely considering the unique homomorphism $\hat{F_2}\to K$ mapping $a\mapsto a'$, $b\mapsto b'$; density implies that it's surjective and hence we can choose $c$ as any preimage of $c$.

Let's now find $K$ (there are many alternative ways).
Consider the three matrices
$$A=\begin{pmatrix}4 & 5\\3 & 4\end{pmatrix},B=\begin{pmatrix}7 & 8\\6 & 7\end{pmatrix},C=\begin{pmatrix}4 & 9\\3 & 7\end{pmatrix}.$$
We'll successively view them as real, and as 3-adic matrices.

These are loxodromic elements in $\mathrm{SL}_2(\mathbf{R})$, with pairwise disjoint pairs of fixed points in the projective line. Hence they have the property that for any large enough $n$, say $n\ge N_0$, the powers $A^n,B^n,C^n$ freely generate a free group.

This was the use of real numbers. Now view them as 3-adic matrices. Observe that all of them belong to the subgroup $K$ of $\mathrm{SL}_2(\mathbf{Z}_3)$ of those matrices $\begin{pmatrix}x & y\\z & t\end{pmatrix}$ such that $x-1,t-1,z$ are in $3\mathbf{Z}_3$ (i.e. the inverse image of the upper unipotent subgroup in reduction modulo 3).

The profinite group $K$ is actually a pro-$3$-group. Its elementary abelianization is given by the homomorphism $f:\begin{pmatrix}1+3a & b\\3c & 1+3d\end{pmatrix}\mapsto (b,c)\in (\mathbf{Z}/3\mathbf{Z})^2=:L$. A basic fact in pro-$p$-groups is that a subgroup $\Gamma$ of $K$ is dense iff $f(\Gamma)=L$. Since $f(A)=(1,2)$, $f(B)=(2,2)$ and $f(C)=(1,0)$ which are pairwise non-collinear, this applies to the subgroup generated by any of the three pairs in $\{A,B,C\}$.

Moreover, for any $n$ with $n$ coprime to 3 and any $u\in\{A,B,C\}$, the cyclic subgroups generated by $u$ and $u^n$ have the same closure in $K$. If furthermore $n\ge N_0$, we deduce that $\langle A^n,B^n\rangle$ is still dense and $(A^n,B^n,C^n)$ freely generates a free subgroup. So this is the desired pair $(a',b',c')$.