Let $R$ be a complete dvr with fraction field $K$ and residue field $k$, and let $X, Y$ be two smooth projective $R$-schemes with isomorphic generic fibers.

Is it true that $[X_k]=[Y_k]$ in $K_0(\text{Var}_k)$?

Recall that $K_0(\text{Var}_k)$ is the so-called Grothendieck ring of varieties, namely, the free Abelian group on $k$-varieties, modulo the relation $[X]=[Y]+[X\setminus Y]$ for $Y\subset X$ closed.

That's the short version; let me say why I would call this a Cauchy integral formula and say what I know.

First of all, why is this a Cauchy integral formula? In this analogy, I'm thinking of a morphism $X\to Y$ as a function $Y(T)\to K_0(\text{Var}_T)$, sending a point $y$ to $X_y$. You're supposed to think of $[X_K]$ as a function $\text{Spec}(K)$ (a puntured disk), and the question asks if we can recover the value at the central point $\text{Spec}(k)$ from this data. The complex-analytic analogue is exactly the Cauchy integral formula.

Let me give a slightly non-trival example. A trivial family of Hirzebruch surfaces $X_n$ may degenerate to either $X_n$ or $X_{n+2}$. But of course $[X_n]=[X_{n+2}]=(\mathbb{L}+1)^2$.

Some other examples: if $X_K, Y_K$ are isomorphic as polarized varieties and one of them is not ruled, all is good by Matsusaka-Mumford. In particular, if $X, Y$ are canonically polarized, the answer to my question is affirmative.

Likewise suppose $X_K, Y_K$ have trivial canonical bundle. Then $X_k, Y_k$ have trivial canonical bundle as well (at least if $K$ has characteristic zero) and are birationally equivalent, by spreading out the isomorphism $X_K\simeq Y_K$. But birational Calabi-Yau's have the same class in (some mild localization of) $K_0(\text{Var}_k)$, by Kontsevich's motivic integration results.

A slightly better version of the question is:

Let $X, Y$ be smooth projective $R$-schemes with $[X_K]=[Y_K]$. Does this imply $[X_k]=[Y_k]$?

This would give some kind of specialization map $K_0(\text{Var}_K)\to K_0(\text{Var}_k)$, at least if $\text{char}(K)=0$.

ADDED: The result is also true for curves and surfaces. For curves this is easy; for surfaces, observe that $X_k, Y_k$ are birationally equivalent. Thus there is a surface $Z$ obtained by blowing up $X_k, Y_k$ at some number of points, so $$[X_k]+n\mathbb{L}=[Z]=[Y_k]+m\mathbb{L}.$$ But $n=m$ because $X_k, Y_k$ have equal Euler characteristic, so $[X_k]=[Y_k]$.