The expected degree for the Delaunay triangulation on hyperbolic space will depend on the density of the points. With low enough density points, you should get arbitrarily high degree. You should be able to get the expected degree as high as you want for a point distribution in the plane by taking the conformal representation of the hyperbolic plane as a varying metric in a unit disc (the Poincare disc), and placing the points with density proportional to this metric. Then the density goes to infinity at the boundary of the circle.

Note that since the Poincare disc model takes circles in the hyperbolic plane to circles in the disc, the Delaunay triangulation of a set of points (which is characterized by the fact that every triangle does not contain another point) is the same in the disc and the hyperbolic plane. Thus, the expected average degree of the Delaunay triangulation should be the same in both of these scenarios. 

The above construction satisfies the conditions stated in your question, but I don't know if you'll be satisfied. Did you want the points to be distributed on the entire plane? If so, I expect it's impossible because there's no way to put a conformal metric on all of $\mathbb{R}^2$ that gives the hyperbolic plane (although note that this isn't quite a proof).