You are looking for

compact 3-manifolds that admit a hyperbolic metric with geodesic boundary

equivalently

compact 3-manifolds that do not contain any essential surface with $\chi \geq 0$

with the additional requirement that every boundary component has the same genus $g$.

To construct such manifolds you may draw pictures of suffficiently knotted graphs in $S^3$ consisting of some copies of genus-$g$ graphs, and take their complements. Then you can use orb to check whether the complement has a hyperbolic structure with geodesic boundary.

An alternative construction uses ideal triangulations, extending Thurson's original "knotted y" example from his notes. Pick a bunch of tetrahedra and pair their faces so that every edge in the resulting triangulation has valence $> 6$. Then remove an open star at each vertex. Geometrization guarantees that the resulting manifold admits a hyperbolic metric with geodesic boundary (because you can put an angle structure *à la Casson* which excludes any normal surface with $\chi \geq 0$).

For example, you can take $g\geqslant 2$ tetrahedra and pair the faces in such a way that the resulting triangulation consists of one vertex and one edge only (which has thus valence $6g$). The resulting manifold is a hyperbolic 3-manifold with connected genus-$g$ geodesic boundary. Its hyperbolic structure is simply obtained by giving each tetrahedron the structure of a truncated regular hyperbolic tetrahedron with all dihedral angles of angle $\pi/(3g)$. Thurston's knotted y is obtained in this way for $g=2$.

The manifolds constructed in this way are "the simplest ones" among those having a connected genus-$g$ boundary, from different viewpoints: they have smallest volume (as a consequence of a result of Myiamoto) and smallest Matveev complexity: we have investigated these manifolds here. There are many such manifolds because there are many triangulations with one vertex and one edge: their number grow more than exponentially in $g$.