Apparently there is no proof in the literature. Let me provide a proof here, in case it's helpful. Actually, first let's give the set-up, then a precise statement, then the proof. To tell the truth, the set-up takes the most time. So this fact is harder to state than to prove, which may be why it's not so useful in practice. But I think it does give valuable intuition for the boundary map.

The set-up: for a noetherian commutative ring R, let $mod_R$ denote its abelian category of finitely-generated modules. Then for a DVR $A$ with residue field $k=A/\mathfrak{m}$ and fraction field F, the composite

$$mod_k \hookrightarrow mod_A \rightarrow mod_F$$

has a canonical trivialization ($0\simeq M\otimes_A F$ whenever $\mathfrak{m}\cdot M=0$), and Quillen's devissage and localization theorems combine to show that the induced map on classifying space of Q-constructions

$$|Q(mod_k)|\rightarrow |Q(mod_A)| \rightarrow |Q(mod_F)|$$

is a fiber sequence (of pointed spaces -- or of spectra after group-completing w.r.t. $\oplus$). The goal is to describe the "boundary", or "monodromy" map

$$\partial: \Omega |Q(mod_F)|\rightarrow |Q(mod_k)|$$

associated to this fibration.

The description will use a different model for the left-hand side $\Omega |Q(mod_F)|$. Namely, let $\mathcal{C}$ denote the following symmetric monoidal category:

Objects of $\mathcal{C}$ are tuples $(L_0,L_1,V)$ where $V$ is a finite-dimensional $F$-vector space and $L_0,L_1$ are spanning f.g. $A$-submodules of $V$ such that $L_0\subset L_1\subset \mathfrak{m}\cdot L_0$.

Morphisms in $\mathcal{C}$ from $(L_0,L_1,V)$ to $(L_0',L_1',V')$ are isomorphisms $\alpha:V\simeq V'$ such that $\alpha(L_0)\supset L_0'$ and $\alpha(L_1)\subset L_1'$ (so $\alpha$ realizes $L_1/L_0$ as a subquotient of $L_1'/L_0'$).

The symmetric monoidal structure is $\oplus$.

Now consider the composition
$$|\mathcal{C}|\rightarrow |(mod_F)^{\simeq}|\rightarrow \Omega |Q(mod_F)|.$$
(By $(-)^{\simeq}$ we mean underlying groupoid.) Here the first map is induced by the forgetful functor $\mathcal{C}\rightarrow (mod_F)^{\simeq}$, and the second map is the usual inclusion of objects of an exact category as loops in the Q-construction. Then on the one hand it is easy to check that the first map is an equivalence (fixing $V$, the space of choices for $L_0$ is contractible because it filters down to zero; and fixing $V$ and $L_0$, the space of choices for $L_1$ is contractible because it has a maximal element $\mathfrak{m}\cdot L_0$); but on the other hand by the comparison of the group-completion and Q-construction approaches to K-theory (see Algebraic K-theory: II) the second map is a group-completion. Thus the composition is a group-completion, so one way to produce maps $\Omega |Q(mod_F)|\rightarrow |Q(mod_k)|$ (such as our desired $\partial$) is to produce a symmetric monoidal functor $\mathcal{C}\rightarrow Q(mod_k)$, then apply geometric realization and group-complete.

This leads us to the precise statement: let $f:\mathcal{C}\rightarrow Q(mod_k)$ denote the symmetric monoidal functor
$$f(L_0,L_1,V) = L_1/L_0.$$
Then the group-completion of $|f|$, identified via the above with a map $\Omega |Q(mod_F)|\rightarrow |Q(mod_k)|$, gives the boundary map in the localization sequence.

Now for the proof. We will use the following characterization of boundary maps in general. Given a fiber sequence $F\rightarrow E\rightarrow B$ and a map $f:\Omega B\rightarrow F$, to give a homotopy from $f$ to the boundary map is equivalent to giving a nullhomotopy of the composite $\Omega B\overset{f}{\longrightarrow} F\rightarrow E$, plus a homotopy between $id_{\Omega B}$ and the map $\Omega B\rightarrow \Omega B$ gotten by combining the two different nullhomotopies of the composition $\Omega B\overset{f}{\longrightarrow} F\rightarrow E\rightarrow B$. (I'm being careless with signs.)

So let us provide this data in our case. The whole time we will be considering maps of spectra out of the group completion of $|\mathcal{C}|$, so we can and will specify all our data just at the level of $\mathcal{C}$. There are two pieces of data:

We need a nullhomotopy of the composite $\mathcal{C}\overset{f}{\longrightarrow} Q(mod_k)\rightarrow Q(mod_A)$ after geometric realization. This is provided by a composite of two natural transformations, one of which goes the wrong way. The first is the obvious identification of $L_1/L_0$ with a subquotient of $L_1$; the second is the obvious identification of $0$ with a subquotient of $L_1$.

We need a homotopy of the resulting map $|\mathcal{C}|\rightarrow \Omega |Q(mod_F)|$ (gotten by combining the nullhomotopy of 1. with the nullhomotopy provided by our fiber sequence) with the canonical map considered above. This is clear: when we tensor the null-homotopy of 1. with $F$ over $A$ and use $0\simeq (L_1/L_0)\otimes_A F$ and $L_1\otimes_A F\simeq V$, we see exactly the usual way of mapping $|(mod_F)^\simeq|$ to $\Omega |Q(mod_F)|$, the one used to compare group-completion to Q-construction.

That gives the proof.