Let $R$ be a commutative ring. An $R$-module $M$ is torsion-free, if for every regular (=non-zero-divisor) element $r\in R$ and every $m\in M$ it holds that $rm \neq 0$. We say a full subcategory $\mathcal{T}\subseteq\mathsf{Mod}_R$ is thick, if for every short exact sequence $$0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$$ it holds that $M \in \mathcal{T} \iff M',M'' \in \mathcal{T}$, i.e. if it is closed under subobjects, quotients and extensions.
It is easy to see that the full subcategory $\mathsf{Tor}_R \subseteq \mathsf{Mod}_R$ of torsion modules is thick. If we restrict to finitely generated modules over a principal ideal domain, by the structure theorem this is the maximal thick subcategory of $\mathsf{Mod}_R^\text{fg}$, since any torsion-free module is free. This is to say, as soon as a thick subcategory $\mathcal{T}$ contains a torsion-free module, it holds that $\mathcal{T} = \mathsf{Mod}_R^\text{fg}$.
How does this generalize to arbitrary (commutative) rings?
I am aware that over a general ring, torsion-free and free are distinct notions. If in general every torsion-free module $M$ would contain a free submodule $F$, then we would be done by means of the short exact sequence $$0 \rightarrow F \rightarrow M \rightarrow M/F \rightarrow 0.$$ But unfortunately this is not true in general. A counterexample is given by the torsion-free $\mathbb{Z}/4\mathbb{Z}$-module $\mathbb{Z}/2\mathbb{Z}$.
Note however, that this is not a counterexample to the maximality statement. Indeed the short exact sequence of $\mathbb{Z}/4\mathbb{Z}$-modules $$0 \rightarrow 2\mathbb{Z}/4\mathbb{Z} \rightarrow \mathbb{Z}/4\mathbb{Z} \rightarrow \mathbb{Z}/2\mathbb{Z} \rightarrow 0$$ shows that $\mathbb{Z}/4\mathbb{Z}$ is an extension of $\mathbb{Z}/2\mathbb{Z}$ by itself, hence the thick subcategory generated by $\mathbb{Z}/2\mathbb{Z}$ is all of $\mathsf{Mod}_{\mathbb{Z}/4\mathbb{Z}}^\text{fg}$...
So what I want to ask is:
Does the thick subcategory generated by a torsion-free module always contain a free module?
You can turn your example over $\mathbb{Z}/4\mathbb{Z}$ into an actual counterexample by making it more "infinite" so that you can't build a free module out of finitely many extensions of your module. For instance, let $R=k\oplus V$ where $k$ is a field, $V$ is an infinite-dimensional $k$-vector space, and you make $R$ a ring by saying $xy=0$ for all $x,y\in V$. Then the $R$-module $k=R/V$ is torsion-free, but clearly the thick subcategory it generates contains no nontrivial free module since it only contains modules that are finite-dimensional over $k$.
Or, if you want a Noetherian example, you could take $R=k[x,y]/(xy)$ and $M=R/(y)$. Then every module in the thick subcategory generated by $M$ is annihilated by some power of $y$, and thus cannot be a nontrivial free module.