It is true, and your idea works (we lift the obstruction).
Proposition. Let $R$ be a ring and let $M$ be a finitely presented $R$-module. If $f : M \to M$ is an automorphism, then there is an epimorphism $\pi : G \to M$ from a finite free $R$-module $G$ and an automorphism $\widetilde{f~} : G \to G$ such that
$$\begin{array}{ccc} G & \xrightarrow{\widetilde{~f~}} & G \\ \pi \downarrow ~~ && ~~\downarrow \pi \\ M & \xrightarrow{f} & M \end{array}$$
commutes.
Partial proof. Choose a presentation
$$K \xrightarrow{i} F \xrightarrow{p} M \to 0$$
where $K,F$ are finite free. We will prove that $f$ lifts to an automorphism of $G := F \oplus K$.
In the following, I will also assume that $i$ is a monomorphism. I am pretty sure that the proof can be fixed without doing that assumption.
The rest of the proof works in an arbitrary abelian category, we will just need the presentation and that $K,F$ are projective.
Let $g := f^{-1}$. Let $K' := \mathrm{im}(i)= \ker(p)$. Since $F$ is projective, there are homomorphisms $\overline{f},\overline{g} : F \to F$ with
$$p \overline{f} = fp, \quad p \overline{g} = g p.$$
The homomorphisms $\overline{f}, \overline{g} : F \to F$ don't have to be inverse to each other, but we can calculate the obstruction: We compute
$$p(\overline{f} \overline{g}-1) = p\overline{f} \overline{g}-p = fp \overline{g} - p = f g p - p = 0.$$
Thus, $\overline{f} \overline{g}-1$ lands inside $K'$. Since $F$ is projective and $K \to K'$ is an epimorphism, there is a homomorphism $\alpha : F \to K$ with
$$\overline{f} \overline{g} - 1 = i \alpha.$$
Similarly, there is a homomorphism $\beta : F \to K$ with
$$\overline{g} \overline{f} - 1 = i \beta.$$
Next, since $p \overline{f} i = f p i = 0$ and $K$ is projective, there is a homomorphism $\gamma : K \to K$ with
$$\overline{f} i = i \gamma.$$
Similarly, there is a homomorphism $\delta : K \to K$ with
$$\overline{g} i = i \delta.$$
Recall that endomorphisms of a direct sum can be represented as matrices:
$$\mathrm{End}(F \oplus K) = \begin{pmatrix} \mathrm{Hom}(F,F) & \mathrm{Hom}(K,F) \\ \mathrm{Hom}(F,K) & \mathrm{Hom}(K,K) \end{pmatrix}$$
Also, composition of endomorphisms is the usual matrix multiplication.
We consider the endomorphisms
$$\widetilde{f~} := \begin{pmatrix} \overline{f} & -i \\ \beta & -\delta \end{pmatrix}, \quad \widetilde{g~} := \begin{pmatrix} \overline{g} & -i \\ \alpha & -\gamma \end{pmatrix}.$$
Let us compute the products.
$$\widetilde{f~} \widetilde{g~} = \begin{pmatrix}
\overline{f} \overline{g} - i \alpha & -\overline{f} i + i \gamma \\
\beta \overline{g} - \delta \alpha & -\beta i + \delta \gamma
\end{pmatrix}$$
The entries in the first row are $1,0$ by construction. To compute the other entries, we use the assumption that $i$ is a monomorphism. Then we may compose with $i$. So to prove $\beta \overline{g} - \delta \alpha = 0$, it suffices to compute
$$i(\beta \overline{g} - \delta \alpha)=i \beta \overline{g} - i \delta \alpha = (\overline{g} \overline{f}-1) \overline{g} - \overline{g} i \alpha=(\overline{g } \overline{f}-1)\overline{g} - \overline{g}(\overline{f} \overline{g}-1) = 0.$$
And to prove $-\beta i + \delta \gamma = 1$, it suffices to compute
$$i(-\beta i + \delta \gamma) = -(\overline{g} \overline{f}-1)i + (\overline{g}i)\gamma = -(\overline{g} \overline{f}-1)i + \overline{g}(\overline{f}i) = i.$$
Thus,
$$\widetilde{f~} \widetilde{g~} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = 1.$$
The same proof shows $\widetilde{g~} \widetilde{f~} = 1$. Thus, $\widetilde{f~}$ is an automorphism of $F \oplus K$.
For $\mathrm{pr}_F : F \oplus K \to F$ we have
$$\mathrm{pr}_F \widetilde{f~} = \overline{f} \, \mathrm{pr}_F - i \, \mathrm{pr}_K$$
by construction of $\widetilde{f~}$. Hence, the composition $\pi = p \circ \mathrm{pr}_F : F \oplus K \to M$ now satisfies
$$\pi \widetilde{f~} = p \, \mathrm{pr}_F \widetilde{f~} = p (\overline{f} \, \mathrm{pr}_F - i \, \mathrm{pr}_K) = p \overline{f} \, \mathrm{pr}_F = f p \, \mathrm{pr}_F = f \pi,$$
and we are finally done.
