To give a more simple example than Daniel's, you can just consider for X a projective line with a single orbifold point. By Riemann-Hurwitz X is simply connected and so there is no non-trivial finite étale morphism Y→X. This holds over an algebraically closed field of characteristic zero say (but would work in characteristic p as well by defining precisely X as a stack of roots in the sense of Vistoli - see Charles Cadman, Using stacks to impose tangency conditions on curves, for the precise definition).

Also, you may want to consider the following closely related notion, taken from

Fundamental Groups of Algebraic Stacks Behrang Noohi http://arxiv.org/abs/math/0201021

"An algebraic stack being uniformizable means that it has a finite étale representable cover by an algebraic space (roughly speaking, its “universal cover” is an algebraic space)."

The author proceeds to show that, roughly, a DM stack X is uniformizable iff all morphisms from the stabilizers to the fundamental group of X are injective.

anyatlas (finite etale or just etale surjective) of $X$ is hyperbolic. In the case of orbifold curves this certainly implies that X is hyperbolic, as an orbifold curve with a hyperbolic finite etale atlas has universal covering $\mathbb H$. $\endgroup$nota counterexample. I do not give any counterexample (I proposed an example, but it turns out not to be a counterexample). $\endgroup$1more comment