Let $S=S_{g,n}$ be a connected compact oriented hyperbolic surface of genus $g$ and $n$ boundary components. Assume that the topological complexity $\xi(S)=3g-3+n$ is at least $2$.
Suppose $p:X\to S$ is a finite regular covering space of $S$. An essential subsurface $Y\subset S$ is a subsurface whose boundary curves are either essential in $S$ or contained in $\partial S$.
Is there an essential subsurface $Y\subset S$ containing $\partial S$such that its preimage $p^{-1}(Y)$ is disconnected?
In the case where the number of boundary components of $X$ and $S$ are equal, this is not possible since the boundary of $X$ is the lift of the boundary of $S$. In particular, any deck transformation taking one component of $p^{-1}(Y)$ to another necessarily takes some component of $\partial X$ to another component of $\partial X$, both of which project to the same component in $\partial S$. This contradicts the assumption that $|\partial X| = |\partial S|$.
On the other hand, one can envision a case where $Y$ is disjoint from some non-separating curve $\alpha$ and the cover is constructed by gluing copies of $S-\alpha$ along $\alpha$. In this case, clearly $p^{-1}(Y)$ is disconnected and the number of components in $\partial X$ is equal to the number of components in $\partial S$ times the degree of the cover.
Having considered these two cases, I suspect that there is a $Y \supset \partial S$ with $p^{-1}(Y)$ disconnected if and only if the preimage $p^{-1}(B)$ of every boundary component $B\subset \partial S$ is disconnected. One direction follows similarly from the $|\partial S|=|\partial X|$ case. I am having difficulties proving/coming up with a counter example for the other direction:
If $p^{-1}(B)$ is disconnected for every component $B\subset \partial S$, can one construct a subsurface $Y\supset \partial S$ such that $p^{-1}(Y)$ is disconnected?