In the same vein as dodd's answer, a counterexample can also be deduced from the second Houghton group $H_2$, which is defined as the group of bijections $L^{(0)} \to L^{(0)}$ that preserves adjacency and non-adjacency for all but finitely pairs of vertices in the bi-infinite line $L$. A presentation of $H_2$ is
$$\left\langle \sigma_i (i \in \mathbb{Z}), t \left| \array{ \sigma_i^2=1, \ i \in \mathbb{Z} \\ [\sigma_i,\sigma_j]=1, \ |i-j| \geq 2}, \ \array{\sigma_i\sigma_{i+1}\sigma_i= \sigma_{i+1}\sigma_i \sigma_{i+1} = 1, \ i \in \mathbb{Z} \\ t\sigma_it^{-1}= \sigma_{i+1}, \ i \in \mathbb{Z}} \right. \right\rangle$$
where $t$ corresponds to a unit translation and $\sigma_i$ to the permutation $(i,i+1)$. Now, truncate the presentation and define $G_n$ via
$$\left\langle \sigma_i (i \in \mathbb{Z}), t \left| \array{ \sigma_i^2=1, \ i \in \mathbb{Z} \\ [\sigma_i,\sigma_j]=1, \ n \geq |i-j| \geq 2}, \ \array{\sigma_i\sigma_{i+1}\sigma_i= \sigma_{i+1}\sigma_i \sigma_{i+1} = 1, \ i \in \mathbb{Z} \\ t\sigma_it^{-1}= \sigma_{i+1}, \ i \in \mathbb{Z}} \right. \right\rangle.$$
By using the relations $t\sigma_it^{-1}=\sigma_{i+1}$ in order to remove the generators $\sigma_0,\sigma_{-1},\ldots$ and $\sigma_{n+2},\sigma_{n+3},\ldots$, we find the following presentation of $G_n$:
$$\left\langle \sigma_1, \ldots, \sigma_{n+1}, t \left| \array{ \sigma_i^2=1, \ 1 \leq i \leq n+1 \\ [\sigma_i,\sigma_j]=1, \ |i-j| \geq 2}, \ \array{\sigma_i\sigma_{i+1}\sigma_i= \sigma_{i+1}\sigma_i \sigma_{i+1} = 1, \ 1 \leq i \leq n \\ t\sigma_it^{-1}= \sigma_{i+1}, \ 1 \leq i \leq n} \right. \right\rangle.$$
Observe from this presentation that $G_n$ decomposes as an HNN extension of
$$\left\langle \sigma_1,\ldots, \sigma_{n+1} \left| \array{ \sigma_i^2=1, \ 1 \leq i \leq n+1 \\ [\sigma_i,\sigma_j]=1, \ |i-j| \geq 2}, \ \sigma_i\sigma_{i+1}\sigma_i= \sigma_{i+1}\sigma_i \sigma_{i+1} = 1, \ 1 \leq i \leq n \right. \right\rangle,$$
which turns out to be isomorphic to the symmetric group $\mathfrak{S}_{n+2}$, where the stable letter conjugates $\langle \sigma_1,\ldots, \sigma_n \rangle$ to $\langle \sigma_2, \ldots, \sigma_{n+1} \rangle$. Thus, as the HNN extension of a finite group, $G_n$ must be virtually free.

The conclusion is that the canonical quotient maps $G_1 \twoheadrightarrow G_2 \twoheadrightarrow \cdots$ defines a sequence of epimorphisms between virtually free groups that does not stabilise.

**Remark:** By reproducing the above argument almost word for word with the lamplighter group $\mathbb{Z}_2 \wr \mathbb{Z}$ instead of the Houghton group $H_2$ provides the same conclusion. The reason is that these groups have a similar structure: they are of the form $C \rtimes \mathbb{Z}$ for some locally finite Coxeter group $C$ where $\mathbb{Z}$ acts on $C$ via an isometry of the graph defining $C$. (Loosely speaking, all the other groups of this form can be recovered from $\mathbb{Z}_2 \wr \mathbb{Z}$ and $H_2$, so there is no other interesting examples in this direction.)