In a covering $X\to S$ of hyperbolic surfaces with boundary, is there a subsurface $Y\supset \partial S$ whose preimage is disconnected? 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?

 A: This answer is an expanded version of my comments. To make your question meaningful, assume that $X, Y$ are connected and $p: X\to S$ is a nontrivial finite regular covering. 


*

*First of all, the basic covering theory shows that for a connected subsurface $Y\subset S$ the following are equivalent:


(a) $p^{-1}$ is connected. 
(b) $\pi_1(Y,y)$ maps onto the quotient group $G= \pi_1(S)/\pi_1(X)$ under the natural homomorphisms
$$
\phi: \pi_1(Y,y)\to \pi_1(S,y)\to G
$$
where the latter homomorphism is given by the regular covering $X\to S$. 


*Suppose that $S$ is closed. I claim that a subsurface $Y$ always exists. If $G$ is abelian (more generally, non-perfect), then find a simple essential nul-homologous loop $\alpha\subset S$ and let $Y$ be its regular neighborhood. Then $H=\phi(\pi_1(Y,y))$ is trivial and, hence, different from $G$. If $G$ is nonabelian, then $H\ne G$ since $H$ is abelian and $G$ is not. 

*My third remark relates to a version of your question in the case when $S$ is closed. One can ask for an essential subsurface $Y$ which is not an annulus. I do not know how to prove the existence of such $Y$ in general, here is a partial result:
Theorem. Suppose that $G$ is simple (nonabelian). Then there exists a number $L_G$ such that if genus of $S$ is $\ge L_G$ then there exists a non-annular compact connected essential subsurface $Y\subset S$ with  $\phi(\pi_1(Y,y))\ne G$. 
Proof. We will need a deep theorem due to Nat Dunfield and Bill Thurston:
N. Dunfield, W. Thurston, Finite covers of random 3-manifolds. Invent. Math. 166 (2006), no. 3, 457–521. 
They proved the following. Given a group $G$ and a surface $S$ and a homomorphism $\pi_1(S)\to G$, define the Euler class $e(\phi)$ as the image of $H_2(S)$ under the natural homomorphism $\phi_*: H_2(S)\to H_2(G)$. For each $e\in H_2(G)$ define the subset $Epi_e(\pi_1(S), G)$ consisting of epimorphisms $\phi$ with $e(\phi)=e$.   
Dunfield and Thurston proved that there exists $k_G$ such that if $S$ has genus $\ge k_G$ then the automorphism group of $\pi_1(S)$ acts transitively on each set $Epi_e(\pi_1(S), G)$. 
Since $G$ is finite, the group $H_2(G)$ is also finite. Each 2-nd homology class $c$ of $G$ is represented by a map of some closed oriented surface $\Sigma$ to the classifying space $K(G,1)$. Let $r_c$ denote the minimal genus of such surface $\Sigma$ and let $r_G$ denote the maximum of $r_c$'s taken over all elements of $H_2(G)$. 
Define $L_G$ to be the maximum of $k_G, r_G+1$ and $2$ (the latter is the minimal number of generators of $G$). 
Suppose now that genus of $S$ is larger than $L_G$. Then each element $e\in H_2(G)$ can be represented by an epimorphism $\psi: \pi_1(S)\to G$ which factors through a homomorphism $\pi_1(S)\to \pi_1(\Sigma)$ induced by a degree one map $h: S\to \Sigma$ with genus of $\Sigma$ strictly less than that of $S$. (The degree one map $h$ is obtained by collapsing a compact essential subsurface $Z$ of $S$ to a point. The subsurface $Z$ has a single boundary component. In particular, $Z$ is not an annulus.) Thus, $\psi(\pi_1(Z,z))=1\in G$. On the other hand, by the Dunfield-Thurston theorem, every epimorphism $\phi: \pi_1(S)\to G$ can be obtained as a composition $\psi\circ \theta$, where $\theta$ is an automorphism of $S$ induced by a homeomorphism $f: S\to S$. (The latter exists by a Dehn-Nielsen-Baer theorem.) 
Then  taking $Y:= f^{-1}(Z)$, we obtain an essential non-annular compact subsurface of $S$ such that $\phi(\pi_1(Y))=\{1\}\ne G$. qed. 


*Lastly, if $S$ has at least two boundary components and $G$ is cyclic, your question has negative answer. Namely, let $C$ be one of the boundary components. Since $C$ is not nul-homologous in $S$, it represents a generator of $H_1(S; G)$. Hence, there exists a homomorphism $\phi: \pi_1(S)\to G$ which sends $\pi_1(C)$ onto $G$. Therefore, for every connected subsurface $Y$ in $S$ containing $C$, we also have $\phi(\pi_1(Y))=G$. The interesting question is what happens if $G$ is simple nonabelian. 

