IMHO: I think it depends on particular dipole and stereochemistry. However, despite the fact that the molecule of imidazole has a very strong dipole, most of the (-) charge is resonated on the non-H nitrogen. The N-H bond is much less polar than the O-H bond in water, due to lone-pair of the H-bonded nitrogen participating in the aromaticity. Additionally, the vector of the bond's dipole points in the center of the imidazole ring.
So, I would venture a guess that in terms of the fraction of H-bonds, this one will be small. Most will be H-bonds from NH2 and COOH.
[Edit] but I am a lawyer not a chem-major.