Nuclear spin multiplicity of radicals
The short answer is that the nuclear spin and electronic spin are separate effects. So an electronic radical ($\ce{N2H+}$) does not affect the nuclear spin of the protons or the nitrogen ($\ce{^14N}$ has nuclear spin 1).

So on a basic level, the net charge does not affect the overall nuclear spins.

The somewhat longer answer is that the nuclear spin states depend on the local chemical environment. This means that we can "see" different proton nuclear spins based on the molecular structure. 

If there's are unpaired electrons, paramagnetic NMR requires understanding the coupling of the electron spin and the nuclear spin. It's possible, but not as straight-forward as typical diamagnetic NMR (for example $\ce{^1H}$ peaks may spread over 200-250 ppm, unlike ~8-12 ppm for normal protons).

Two main effects in paramagnetic NMR are pseudo-contact and contact shifts.


  The effect of the contact term arises from transfer of spin polarization to the observed nucleus. Spin polarization is a consequence of the very strong electron-nuclear (NMR detected) interaction. This coupling, also known by EPR spectroscopists as hyperfine coupling, is on the order of MHz, vs the usual internuclear (J) coupling observed in conventional NMR spectra, which are on the order of a few Hz. This difference reflects the large magnetic moment of an electron (−1.00 μB), which is much greater than any nuclear magnetic moment (e.g. for $\ce{^1H}$: 1.52×10−3 μB).