As user61789 points out, there are two natural questions related to this post. First, under the given hypotheses on the (proper) morphism $\pi_X:X\to \text{Spec}(R)$, and assuming that the generic fiber is stable (in the sense of Deligne, Mumford, Mayer, Grothendieck, Knudsen, ...), does there exist a stable model $\pi_Y:Y\to \text{Spec}(R)$ of the generic fiber. Second, assuming that this stable model does exist, then does there exist an $R$-morphism $\nu:X\to Y$ that is an isomorphism on the generic fiber?

I claim that both of these questions have positive answers. I am certain there are references, and it would be best to give one. However, the basic idea (for at least one proof) is that it suffices to produce the invertible sheaf $\mathcal{L}$ that would be $\nu^* \omega_{Y/R}^{\otimes d}$ for $d$ sufficiently large and divisible, assuming that $Y$ and $\nu$ actually exist. For then $Y$ would be the (relative) Proj (over $R$) of the corresponding graded ring of global sections of powers $\mathcal{L}^{\otimes n}$ for $n\geq 0$. Of course it is necessary to prove that this ring is actually a finitely generated $R$-module, that $\mathcal{L}$ is globally generated (relative to $R$), and that the morphism $\nu$ contracts the unstable rational curves. However, all of these things can be checked after base change, e.g., to geometric fibers.

So why should such an invertible sheaf $\mathcal{L}$ exist? Of course there are constructions (and I am sure they are written up in the references). But if you believe that a stable model $Y\to \text{Spec}(R)$ and a morphism $\nu$ exists fppf locally, then you should believe that $\mathcal{L}$ exists already over $\text{Spec}(R)$. That is because the obstruction to descending an invertible sheaf is an element in the Brauer group (interpreted broadly, soas to include the second group cohomology of any automorphism groups of the stable models of fibers), and this is a torsion group. Thus, up to passing from the original invertible sheaf to some sufficiently divisible, positive tensor power, the sheaf will descend. We gave ourselves that freedom through the integer $d$.

Also, since we are working étale locally, we may assume that the family is pulled back from a "versal" family. Then it suffices to produce the invertible sheaf for this versal family. This addresses the issue user61789 raises: what if $X$ is not regular? For a versal family, $X$ is regular, the generic fiber is smooth, fibers with $\geq r$ nodes occur in codimension $\geq r$ over $\text{Spec}(R)$, etc. Also, now the invertible sheaf $\mathcal{L}$ can be constructed on an open subset whose complement is codimension $\geq 2$: via Hartog's theorem / Riemann extension / the S2 property of $X$, this invertible sheaf will extend uniquely to all of $X$. Of course in codimension $1$ in $\text{Spec}(R)$, when there is at most one node on the "special fibers", it is easy to construct $\mathcal{L}$; namely, define $\mathcal{L}$ to be the relative dualizing sheaf $\omega_{X/R}$ twisted down by the (reduced) Cartier divisor in $X$ that is the union of all rational tails of fibers.

This does not mean that we are done! Why should the invertible sheaf $\mathcal{L}$ defined in this way have all the necessary properties? However, if you believe that such an invertible sheaf can be constructed from a stable model $Y$ and morphism $\nu$ after fppf base change, then you must accept that the invertible sheaf defined above (the unique extension of the invertible sheaf defined in codimension $1$) does have these properties.

So why should a stable model $Y$ and a morphism $\nu$ exist fppf locally? I am not going to reprove the existence of a proper Deligne-Mumford stack of stable curves -- that is delicate to prove (particularly in mixed characteristic, although the Artin-Winters proof does simplify the original argument). However, if you believe in the existence of such a proper Deligne-Mumford stack, then, after allowing a base change, you should believe that such a projective model $Y$ and morphism $\nu$ can be obtained after a blowing up of $\text{Spec}(R)$. Using Zariski's Main Theorem, to prove that the blowing up is not necessary, it suffices to prove that for any positive-dimensional fiber of the blowing up, the corresponding family of stable curves is actually Zariski locally trivial. Since all of these stable curves are obtained by blowing down rational fibers of the corresponding at-worst-nodal fiber of $\pi_X$, this is obvious. So, if you believe in the existence of a proper Deligne-Mumford stack, you should believe in the existence of $\mathcal{L}$.

Of course maybe you are trying to go through the proof of the existence of the proper Deligne-Mumford stack of stable curves! In that case, the logic above is pretty much backwards: you construct the stable model from $\mathcal{L}$ and not the other way around. However, my impression is that the question is really about why it is unnecessary to make a base change to construct $Y$ and $\nu$, assuming that a stable model exists after the base change, rather than about how one proves that always a stable model exists after a base change.