Heuristically, I want to know, given a smooth, projective morphism from a scheme to a discrete valuation ring, if the generic fiber can be 'covered' by a family of geometrically integral curves, is it true that the specialization of a general member of this family is also geometrically integral. More precisely,


Let $\pi: \mathcal{X} \to \mathrm{Spec}(R)$ be a smooth, projective morphism of relative dimension at least $2$ and $R$ is a discrete valuation ring with residue field isomorphic to $\mathbb{C}$.
Let $B$ be an irreducible variety mapping surjectively to $\mathrm{Spec}(R)$.
Consider a family of curves $\mathcal{C} \subset \mathcal{X} \times\_R B$ parameterized by $B$ i.e., $\mathcal{C}$ is a closed subscheme in $\mathcal{X} \times\_R B$ and the natural morphism from $\mathcal{C}$ to $B$ is flat and projective. Suppose that the natural morphism from $\mathcal{C}$ to $\mathcal{X}$ is surjective.


Denote by $K$ the fraction field of $R$ and $\mathcal{C}\_K:= \mathcal{C} \times\_R \mathrm{Spec}(K)$ the family of curves parameterized by the generic fiber $B \times\_R \mathrm{Spec}(K)$.
Similarly, denote by $\mathcal{C}\_k:= \mathcal{C} \times\_R \mathrm{Spec}(k)$ the family of curves parameterized by the special fiber $B \times\_R \mathrm{Spec}(k)$. Note that, there are natural flat, projective morphisms from $\mathcal{C}\_K$ (resp. $\mathcal{C}\_k$) to $B\_K$ (resp. $B\_k$).


Is it true that if a general fiber of the natural morphism from $\mathcal{C}\_K$ to $B\_K$ is geometrically integral, then a general fiber for the morphism from $\mathcal{C}\_k$ to $B\_k$ is also geometrically integral?


Any hint/reference will be most welcome.



Degeneration of curves in smooth families
I am just writing my comments as one answer. Without further hypotheses, there are counterexamples. Even without a specific example of $\mathcal{X}$, there are plenty of examples of a $K$-scheme $B\_K$ and a family of smooth, projective, geometrically connected relative curves $\mathcal{C}\_K\to B\_K$ such that for every fppf $R$-scheme $B\_R$ that has $B\_K$ as its generic fiber, for every flat, proper relative curve $\mathcal{C}\_R\to B\_R$ that extends the $K$-family, the geometric fibers over $B\_k$ are not integral.


Indeed, for every proper, flat $R$-scheme $B'\_R$ whose geometric fibers are integral, for every proper, flat morphism $\mathcal{C}'\_R\to B'\_R$ whose fibers are at-worst-nodal, connected curves with ample dualizing sheaf, for every "modification" of the family over a proper, flat $R$-scheme $B\_R$ whose geometric fibers are integral,
$\mathcal{C}\_R \to B\_R$, the "stabilization" of the geometric generic fiber of $\mathcal{C}\_k\to B\_k$ equals the geometric generic fiber of $\mathcal{C}'\_k\to B'\_k$: this is part of the uniqueness in "stable reduction". So if the geometric generic fiber of $\mathcal{C}'\_k\to B'\_k$ is reducible, the same is true for the geometric generic fiber of $\mathcal{C}\_k\to B\_k$.


Thus, to get a positive answer, you need to add the hypothesis that for the family $\mathcal{C}\_K\to B\_K$ whose geometric fibers are assumed to be integral, there exists an extension $\mathcal{C}\_R\to B\_R$ over an fppf $R$-scheme $B\_R$ whose geometric generic fiber over $B\_k$ is integral (just as a proper, flat family of abstract curves, with no morphism to $\mathcal{X}\_R$ specified).


Even with this hypothesis, there are still counterexamples, e.g., the example in my comment where $\mathcal{X}\_K$ is a Hirzebruch surface $\mathbb{P}^1\times \mathbb{P}^1$, yet $\mathcal{X}\_k$ is a different Hirzebruch surface, e.g., the minimal (crepant) resolution of a singular quadric cone in $\mathbb{P}^3$. The usual way to deal with this is to change the question slightly: does there exist an extension $\mathcal{C}\_R \to B\_R\times\_{\text{Spec}(R)} \mathcal{X}\_R$ and an irreducible component $\mathcal{C}\_{k,i}$ of $\mathcal{C}\_k$ satisfying all of your conditions. If you begin with a $K$-family of curves $\mathcal{C}\_K\to B\_K\times\_{\text{Spec}(K)} \mathcal{X}\_K$ that is constructed in a "sufficiently general" way, e.g., the family of all complete intersection curves of sufficiently very ample divisors in given linear equivalence classes on $\mathcal{X}\_K$, then this new question has a positive answer if and only if there exists an irreducible component $\mathcal{X}\_{k,i}$ of $\mathcal{X}\_k$ that is geometrically irreducible.


