I have wondered about such lifts myself, and I want to give what I hope is a tantalizing hint of what such lifts may be able to tell us:

The lift you give of $S_{3}$ from $GL_{2}( \mathbb{Z} )$ to $Aut(F_{2})$ also gives an embedding of $C_{3} = A_{3}$ into $Aut(F_{2})$. Why is this useful? It gives a character-free proof of a congruence about conjugacy classes of a finite group (here $c(G)$ denotes the number of conjugacy classes of the group $G$):

Theorem. If $G$ is a finite group with $|G|$ not divisible by $3$, then $|G| \equiv c(G) \mod{3}$. $\\$
Proof, with character theory: $|G|$ is the sum of the squares of the dimensions of the complex irreducible representations of $G$. The number of these is $c(G)$. Since $|G|$ is not a multiple of $3$, these dimensions aren't multiples of $3$. Now reduce modulo $3$ and obtain the congruence.
Proof, without character theory: $|G|(|G| - c(G))$ is the number of non-commuting ordered pairs of elements of $G$. Since $|G|$ is a nonmultiple of $3$, the congruence may be proved by showing that this set has a number of elements which is a multiple of $3$. Now just let the lift of $\tau$ act on it: If $(x,y)$ is a fixed point, then reading the first coordinate gives $x=y$, which trivially implies $xy=yx$. So the action is fixed-point-free, and we are done.

Theorem. If $G$ is a finite group of odd order, then $|G| \equiv c(G) \mod{8}$.
Proof, with character theory: $|G|$ is the sum of the squares of the dimensions of the complex irreducible representations of $G$. The number of these is $c(G)$. Since $|G|$ is odd, these dimensions are odd. Now reduce modulo $8$ and obtain the congruence.
Proof, without character theory: Instead of lifting $S_{3}$ to $Aut(F_{2})$, lift the dihedral group of order $8$, which is generated by the involutions $(x,y) \to (x^{-1},y)$ and $(x,y) \to (y,x)$. It suffices to check that none of the $5$ involutions of this group has a fixed point in the action on the non-commuting pairs of elements of $G$. This is easy to do, and it involves recalling that, since $|G|$ is odd, $t^{2}=1$ implies $t=1$ for $t \in G$.

This is all well and good, but, it is not the end of the story:

Theorem. The number of rows in the character table of $G$ which are entirely real-valued is the number of conjugacy classes $C$ of $G$ such that $x \in C$ iff $x^{-1} in C$.
Corollary. If $|G|$ is odd, then the only entirely real-valued character of $G$ is the trivial character.

This leads to a strengthening of the character theory argument used above, so that it now proves that $|G| \equiv c(G) \mod{16}$.