I think that, if the curve does not disappear, then the answer is positive.

In fact, let $\pi \colon \mathcal{X} \to \Delta$ your deformation of surfaces over a disk, with $S_1= \pi^{-1}(t_1)$ and $S_2 = \pi^{-1}(t_2)$. The fact that $C_1 \subset S_1$ deforms to $C_2 \subset S_2$ along $\pi$ means that there exists a line bundle $\mathcal{L}$ on the total space $\mathcal{X}$ such that $$\mathcal{L}|_{S_1}=\mathcal{O}_{S_1} (C_1), \quad \mathcal{L}|_{S_2}=\mathcal{O}_{S_2} (C_2).$$

Therefore we have $$(C_1 \cdot C_1) = \mathcal{L} \cdot \mathcal{L} \cdot S_1 = \mathcal{L} \cdot \mathcal{L} \cdot S_2 = (C_2 \cdot C_2),$$ since any two fibers of $\pi \colon \mathcal{X} \to \Delta $ are clearly numerically equivalent.

I think that, if the curve does not disappear, then the answer is positive.

In fact, let $\pi \colon \mathcal{X} \to \Delta$ your deformation of surfaces over a disk, with $S_1= \pi^{-1}(t_1)$ and $S_2 = \pi^{-1}(t_2)$. The assumption that $C_1 \subset S_1$ deforms to $C_2 \subset S_2$ along $\pi$ means that there exists a line bundle $\mathcal{L}$ on the total space $\mathcal{X}$ such that $$\mathcal{L}|_{S_1}=\mathcal{O}_{S_1} (C_1), \quad \mathcal{L}|_{S_2}=\mathcal{O}_{S_2} (C_2).$$

Therefore we have $$(C_1 \cdot C_1) = \mathcal{L} \cdot \mathcal{L} \cdot S_1 = \mathcal{L} \cdot \mathcal{L} \cdot S_2 = (C_2 \cdot C_2),$$ since any two fibers of $\pi \colon \mathcal{X} \to \Delta $ are clearly numerically equivalent.

I think that, if the curve does not disappear, then the answer is positive.

In fact, let $\pi \colon \mathcal{X} \to \Delta$ your deformation of surfaces over a disk, with $S_1= \pi^{-1}(t_1)$ and $S_2 = \pi^{-1}(t_2)$. The fact that $C_1 \subset S_1$ deforms along $\pi$ means that there exists a line bundle $\mathcal{L}$ on the total space $\mathcal{X}$ such that $$\mathcal{L}|_{S_1}=\mathcal{O}_{S_1} (C_1), \quad \mathcal{L}|_{S_2}=\mathcal{O}_{S_2} (C_2).$$

Therefore we have $$(C_1 \cdot C_1) = \mathcal{L} \cdot \mathcal{L} \cdot S_1 = \mathcal{L} \cdot \mathcal{L} \cdot S_2 = (C_2 \cdot C_2),$$ since any two fibers of $\pi \colon \mathcal{X} \to \Delta $ are clearly numerically equivalent.

I think that, if the curve does not disappear, then the answer is positive.

In fact, let $\pi \colon \mathcal{X} \to \Delta$ your deformation of surfaces over a disk, with $S_1= \pi^{-1}(t_1)$ and $S_2 = \pi^{-1}(t_2)$. The fact that $C_1 \subset S_1$ deforms to $C_2 \subset S_2$ along $\pi$ means that there exists a line bundle $\mathcal{L}$ on the total space $\mathcal{X}$ such that $$\mathcal{L}|_{S_1}=\mathcal{O}_{S_1} (C_1), \quad \mathcal{L}|_{S_2}=\mathcal{O}_{S_2} (C_2).$$

Therefore we have $$(C_1 \cdot C_1) = \mathcal{L} \cdot \mathcal{L} \cdot S_1 = \mathcal{L} \cdot \mathcal{L} \cdot S_2 = (C_2 \cdot C_2),$$ since any two fibers of $\pi \colon \mathcal{X} \to \Delta $ are clearly numerically equivalent.