sp² hybridized orbital
The picture below from the UC Davis ChemWiki summarizes why that lone pair is not available. It also explains how pyridine can function as a base.

The lone pair in the $sp^2$ orbital is perpendicular to the $p$ orbitals and $\pi$ bonds in the molecule. the electron pair from the nitrogen that participates in aromaticity is the one in the $\ce{C=N}$ bond.
To answer your first question about hybridization, we have a conservation law that guarantees that when orbitals mix (for any reason), we need to get the same number of orbitals out as we put in. The law is the Conservation of Angular Momentum, which does apply in quantum systems.
Nitrogen, in its valence shell, has four orbitals: $2s,2p_x,2p_y,2p_z$. The $sp^2$ hybridization is achieved by mixing two $p$ orbitals and one $s$ orbital. Because of the conversation of angular momentum, four orbitals in means four orbitals out. Thus, the $p$ orbital that was not mixed is still there (and is the only one still there).
$$[s,p,p],p\implies[sp^2,sp^,sp^2],p$$
We cannot have two $p$ orbitals at an $sp^2$ hybridized atom because then we would have 5 orbitals (or we would not have $sp^2$).

Now, let's create a general rule for aromatic heterocycles.

Look at pyrrole (also from ChemWiki);

In the structure of pyrrole, the $\ce{N-H}$ bond uses the $sp^2$ orbital and the lone pair is in the $p$ orbital. In general, if there is a double bond drawn to the nitrogen atom (like in pyridine), then the $\pi$ bond uses the $p$ orbital and the lone pair is in the $sp^2$ orbital and does not count toward aromaticity. If there is no double bond drawn to the nitrogen atom, usually there is a $\ce{N-H}$ bond (or some other bond to the nitrogen). This $\ce{N-H}$ bond uses the $sp^2$ orbital and the lone pair uses the $p$ orbital and counts toward aromaticity.