**Counterexample**: the 3-ball $B^3$. It is irreducible because it is a compact submanifold of $\mathbb{R}^3$ with connected boundary. It has non-empty boundary and $\partial B$ is simply connected (and incompressible surfaces with boundary must send their boundary to curves which do not bound disks in $\partial B$). Hence if there is an incompressible surface it must be closed. Non orientable closed surfaces do not embed in $B^3$. Moreover, $B^3$ is simply connected and
recall that by Dehn's lemma an orientable, closed surface with $\chi(S)\leq 0$ is incompressible if the inclusion is injective at the level of fundamental group.

However the following is true:

If $M$ is compact with non-empty boundary, oriented , irreducible, and $\partial$-irreducible* then either $M=B^3$ or $M$ contains an incompressible
and $\partial$-incompressible surface.

This result relies on the fact that (under the above assumptions), given a class in $H_2(M,\partial M;\mathbb{Z})$ you can represent it by disjoint union of incompressible and and $\partial$-incompressible surfaces.
*

You can find all the details for example in *Bruno Martelli. An Introduction to Geometric Topology. https://arxiv.org/pdf/1610.02592.pdf*
Proposition 9.4.3 and Corollary 9.4.5.

*Meaning that there are no essential disks, in other words $M$ is not obtained by joining two 3-manifolds with a 1-handle.