How precisely does a star collapse into a black hole?

The solution for this problem for a dust equation of state and spherical symmetry is known as the Oppenheimer-Snyder solution. You model the interior of the distribution as a FRW universe with positive spatial curvature, zero pressure, and zero cosmological constant. You model the exterior of the solution as the Schwarzschild solution cut off at a time-dependent radius. So long as the matter distribution is dust, the thing satisfies all of the junction conditions you need. See Poisson's relativity book or MTW.

A more general solution requires numerics. But one thing we can say for sure is that there is no need for the black hole to shed its 'hair' in the case of spherical symmetry--the radial dependence of the solution will just compress into the singularity eventually, or scatter out to infinity. Birchoff's theorem tells us every spherically symmetric vacuum solution must be the Schwarzschild solution (perhaps with an electrostatic charge, which is technically not vacuum). This is related to the fact that there can be no monopole radiation in relativity.

Also, the general case for this problem is very likely chaotic. Already, if the equation of state of the matter is that of a classical, spherically symmetric, Klein-Gordon field, which is a relatively simple generalization, the system exhibits a (link is a large postscript file)second-order phase transition, a result found by Matt Choptuik, and related to the settling of the Hawking naked singularity bet.

The "hair" is lost via gravitational radiation. This is also known as quasi-normal ringdown, as the BH vibrates at different frequencies much like a drum (maybe a "gong" is better analogy). Any charge on the black hole will simple get shorted out by free charges in the surrounding plasma, on a very short time scale.