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.