Why isn't this resonance possible in an imidazole ring?
You need to understand the basic reason for the aromatic nature of materials. A molecule is said to be aromatic only if:


A ring in it is planar, 
Continuous delocalisation is possible throughout the concerned cyclic ring, and
This delocalisation stabilises the ring.


Now, for continuous delocalisation to be possible, the $\mathrm{\pi}$ orbitals of all the elements in the ring need to be parallel to each other ( They need to overlap sideways for formation of  $\mathrm{\pi}$ bonds).

In your example, consider the carbons in the ring. All of their $\mathrm{\pi}$ orbitals are parallel to each other and PERPENDICULAR to the plane of paper. On the other hand, the imine nitrogen is $\mathrm{sp^2}$ hybrid. The lone pair of electrons is in an  $\mathrm{sp^2}$ hybrid orbital, i.e, IN the plane of the paper. And mind you, it's fixed (can't rotate without distorting the geometrical shape of the ring, that is, which would be too much to expect from the poor guy.) 

So the imine Nitrogen cannot give the lone pair anywhere. it can only play around with the  $\mathrm{\pi}$ bond already on it.

Moreover, as soon as the amine Nitrogen takes up a proton, it loses its ability to use the lone pair for delocalisation. The condition for planarity itself is not fulfilled ($\ce{(R)2-N+-(H)2}$ has tetrahedral geometry), so there's no use of discussing aromaticity. Before the proton uptake, the ring was aromatic because the amine Nitrogen was using the lone pair for continuous delocalisation, effectively attaining planar geometry. But now... It can no longer be done. 

Conclusion: Resonance is still possible(between all atoms of the ring except the amine Nitrogen), but Continuous Delocalisation is not. So, the system loses its aromaticity.