Hi stupid_question_bot,
Unfortunately you seem to need some more assumptions for an easy proof, in particular properness would make this very easy, in general given a flat proper scheme with geometrically normal fibers one can show that the number of (geometric) components of the fibers is locally constant on the base, which would answer your question.
(EDIT: To be clear, the following is not a counter-example to the specific statement in the question, that comes later in this answer. I was just trying to point out that the proof would need some geometric input since it is false when the base is not normal.)
The counter example that I have in mind is as follows, take $\bar{X}$ to be the nodal cubic over $\mathbb{Z}_p$ ($\mathbb{P}^1$ glued together at two $\mathbb{Z}_p$ points: say $0, 1$ in a standard affine chart), let $\bar{Y} \to \bar{X}$ be a connected finite etale cover corresopnding to a nontrivial element of the geometric fundamental group of $\bar{X}$ (for definiteness, take the double cover given by two $\mathbb{P}^1$'s glued into a bigon and for safety let $p \neq 2$). Now let $X$ be the complement of the node in the special fiber, and let $Y$ be the pullback. Clearly while the generic fiber of $Y$ is connected the special fiber is not by inspection.
You can now complain: "oh but your $X$ is not an snc complement in a smooth scheme." In this case I was unable to say anything useful, except for that some results in SGA imply that this would be true if the cover $Y$ is tamely ramified over the snc divisor. Hope this example is helpful though, as it shows that the strong statement you made about connectivity of special fibers is not some total triviality.
EDIT: Update, bad news: there are even worse examples to be had here. Let $X$ be $\mathbb{A}^1_{\mathbb{F}_p[[t]]}$, then consider $Y$ the Artin-Schreier cover of $X$ given by the equation $Y^p - Y = x \cdot t$, then the special fiber of this etale cover splits but generically it defines a Galois Artin-Schreier cover.
