Does deuterium exhibit hydrogen bonding?
(Based on ron's answer here on the inductive effect accorded by deuterium - do give that a read, and consider upvoting it) 

Since deuterium has a higher mass than protium, simple Bohr theory tells us that the deuterium 1s electron will have a smaller orbital radius than the 1s electron orbiting the protium nucleus.
The smaller orbital radius for the deuterium electron translates into a shorter (and stronger) $\ce{C-D}$ bond length.
 See the bottom half of ron's answer for the proof 

In your case, that would mean a shorter $\ce{N/O/F-D}$ bond.

A shorter bond has less volume over which to spread the electron density (of the 1 electron contributed by $\ce{H}$ or $\ce{D}$) resulting in a higher electron density throughout the bond, and, consequently, more electron density at the carbon end of the bond. Therefore, the shorter $\ce{C-D}$ bond will have more electron density around the carbon end of the bond, than the longer $\ce{C-H}$ bond.
The net effect is that the shorter bond with deuterium increases the electron density at carbon, e.g. deuterium is inductively more electron donating than protium towards carbon.

So we can expect the $\ce{N/O/F-D}$ bond to be smaller and more polar than the corresponding bond with protium.
Karl very astutely pointed out that the boiling point of heavy water is higher than that of normal water. In light of the inferences drawn earlier, I took the liberty of interpreting this fact as being (at least somewhat) indicative of deuterium permitting a stronger hydrogen bond than protium.
As Linear Christmas pointed out, the variations in mass across different isotopes is known to affect boiling point (as in the case of helium). This is likely to contribute to some degree, to the increase in boiling point in heavy water. However, I'm still of the opinion that the higher dipole moment of $\ce{D2O}$ over $\ce{H2O}$ is the main cause for the higher boiling point.

Finis
Hydrogen bonding with deutrium does occur and should be, in theory (drawn from ron's answer), stronger than that with protium. The fact that heavy water $\ce{D-O-D}$ is more polar (and hence has a higher boiling point) than regular water $\ce{H-O-H}$, appears to be a consequence of this.