Indeed, it seems that the situation gets nicer, but not as nice as what I depicted in my first and **very flawed** answer (see its remains below and the enlightening counter-example of Wilberd van der Kallen).

The keyword is **torsion-free modules of finite rank** over a discrete valuation ring. You can find a classification (up to $R$-isomorphism and quasi-isomorphism) of the $R$-submodules of $K^n$ for $R$ a DVR and $K$ its fraction field in [1, Corollary 1.7]. The characterization is given via conditions on matrices with coefficients in the completion of $R$ which represent the $R$-modules. Other useful results can be found in [2].

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**Edit**: There is a **flaw** in the proof below. Indeed, I didn't manage to establish that $(3)$ implies $(1)$ but only that $(3)$ implies $(2)$.
The lemma below is simply **wrong**.

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>> **Claim**. Let $R$ be DVR and let $K$ be its fraction field. An $R$-submodule of $K^n$ is isomorphic to an $R$-module the form $K^m \oplus R^k$ for some non-negative integers $m$ and $k$ such that $m + k \le n$.

We will make use of

>> **Lemma.** Let $R$ be a DVR and let $K$ be its fraction field. Let $M \subseteq K^n$ be an $R$-submodule. Then the following are equivalent: 

>>  1. $M$ contains a $K$-subspace of positive dimension.
>>  2. There is a surjective $R$-homomorphism from $M$ onto $K$.
>>  3. $M$ is not finitely generated. 

>> *Proof*. Since a $K$-vector space is an injective $R$-module, hence a direct summand of any of its containing modules, we infer that $(1)$ implies $(2)$. Since $K$ is not finitely generated as an $R$-module, we also infer that $(2)$ implies $(3)$. Assume no that $M$ is not finitely generated. Then some of its images under the projection maps $K^n \twoheadrightarrow K$ must be $K$ since otherwise $M$ would embed into $R^n$. Therefore $(1)$ holds. 

We are now in position to prove the claim.

>> *Proof of the claim*. Let $M$ be a submodule of $K^n$ and let $m \ge 0$ be the largest integer such that $M$ contains a $K$-subspace of $K^n$ of dimension $m$. Since $K^m$ is an injective $R$-module, it is a direct summand of $M$. Therefore we can write $M = K^m \oplus N$ for some $R$-submodule $N$ of $K^n$. As $N$ doesn't contain any $R$-submodule isomorphic to $K$, it is finitely generated by the above lemma. Because $N$ is torsion free, it is free of finite rank.

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[1] D. Arnold, "A duality Torsion-free modules of finite rank over a discrete valuation ring", 1969.  

[2] D. Arnold, M. Dugas and K. Rangaswamy, "Torsion-free modules of finite rank over a discrete valuation ring", 2002.