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, and placing the points with density proportional to this metric. Then the density goes to infinity at the boundary of the circle.

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.