I suspect there is a bit of a difficulty here. A particle $p$ and its anti-particle $\bar p$ cancel each other, and if they have a mass $m$ this results in the production of massless bosons (photons). Quantum number which identify the particle $p$ are subtracted away by opposite quantum numbers of $\bar p$. These quantum numbers usually are the lepton or baryon numbers, electric charge, isospin and so forth. However, anti-matter does not have anti-mass.
Dirac’s original idea was that the Klein-Gordon equation had a square-root according to spinors. If the particle has a mass there are then two roots for the momentum, which define a surface in $\pm$ portions of the momentum light cone. The negative portion of the cone defines the so called Dirac sea, where the mass of particles is negative. All of these negative energy (mass) states are however completely filled up. This is why this is called a “sea,” for they define a ground state which has no dynamics. However, if you impart a package of energy to a state in the sea so its energy flips sign you can generate an anti-particle with positive mass-energy. However, all the other quantum numbers are reversed, including the charge.
One could then construct a black hole from a huge cloud of hydrogen and anti-hydrogen of equal mass. Based on the final state of the black hole it is not possible to determine whether it was formed by hydrogen or anti-hydrogen. So if you have two such black holes, one from $H$ and the other from $\bar H$ the two will coalesce into a larger black hole.
The Schwarzschild solution in its pure form has a past and future singularity, where the past singularity corresponds to a “white hole.” The white hole is about the closest thing there is to an “anti-black hole.” These do not exist in nature, or at least have not been identified astronomically. They may play some role in the very early universe, but nature is such that there is an asymmetry (asymmetry in their occurrence etc) between the black hole and white hole. However, the white hole does not have an “anti-mass.”
You need to be a lot more careful when you use the phrase red-shift, due to how frequency is measured in general relativity. Roughly speaking, a photon is characterised by its wave vector $k$, which is a light-like four vector. The frequency measured by an observer is $g(\tau,k)$ where $g$ is the metric tensor, and $\tau$ is the unit vector tangent to the observer's world-line.
A little bit of basic Lorentzian geometry tells you that for any given photon $k$, instantaneously there can be observers seeing that photon with arbitrary high and arbitrarily low frequency.
So: start with your space-time. Fix a point on the event horizon. Fix a photon passing through that space-time event. For any frequency you want to see, you can choose a time-like vector at that space-time event that realises that frequency. Now, since the vector is time-like and the event horizon is null, the geodesic generated by that vector must start from outside the event horizon and crosses inside. Being a geodesic, it represents a free fall. So the conclusion is:
For any frequency you want to see, you can find a free falling observer starting outside of the black hole, such that it crosses the event horizon at the given space-time event and observes the frequency you want him to see.
So you ask, what is this whole business about gravitational red-shift of Schwarzschild black holes? I wrote a longer blog post on this topic some time ago and I won't be as detailed here. But the point is that on the Schwarzschild black holes (and in general, on any spherically symmetric solution of the Einstein's equations), one can break the freedom given by local Lorentz invariance by using the global geometry.
On Schwarzschild we have that the solution is stationary. Hence we can use the time-like Killing vector field* for the time-translation symmetry as a "global ruler" with respect to which to measure the frequency of photons. This is what it is meant by "gravitational redshift" in most textbooks on general relativity (see, e.g. Wald). Note that since we fixed a background ruler, the frequency that is being talked about is different from the frequency "as seen by an arbitrary infalling observer".
(There is another sense in which redshift is often talked about, which involves two infalling observers, one "departing first" with the second "to follow". In this case you again need the time-translation symmetry to make sense of the statement that the second observer "departed from the same spatial point as the first observer, but at a later time.)
It turns out, for general spherically symmetric solutions, there is this thing called a Kodama vector field, which happens to coincide with the Killing vector field on Schwarzschild. Outside of the event horizon, the Kodama vector field is time-like, and hence can be used as a substitute for the global ruler with respect to which to measure red-shift, when the space-time is assumed to be spherically symmetric, but not necessarily stationary. Again, this notion of redshift is observer independent. And it has played important roles (though sometimes manifesting in ways that are not immediately apparently related to red-shift, through choices of coordinates and what-not) in the study of dynamical, spherically symmetric gravitational collapse in the mathematical physics literature.
To summarise:
If you just compare the frequency of light measured (a) at its emission at the surface of the star in the rest frame associated to the collapse and (b) by an arbitrary free-falling observer, you can get basically any values you want. (Basically because the Doppler effect depends on the velocity of the observer, and you can change that to anything you like by choosing appropriate initial data for the free fall.)
One last comment about your last question:
You asked about what happens in the interior of the black hole. Again, any frequency can be realised by time-like observers locally. The question then boils down to whether you can construct such time-like observers to have come from free fall starting outside the black hole. By basic causality considerations, if you start with a time-like vector at a space-time event inside the black hole formed from gravitational collapse, going backwards along the time-like geodesic generated by the vector you will either hit the surface of your star, or exit the black hole. Though precisely how the two are divided depends on the precise nature of the gravitational collapse.
I should add that if you use the "global ruler" point of view, arguments have been put forth that analogous to how one expects red shift near the event horizon, one should also expect blue shifts near any Cauchy horizon that should exist. This has been demonstrated (mathematically) in the Reissner-Nordstrom (and similar) black holes. But as even the red-shift can sometimes run into problems (extreme charged black holes), one should not expect the statement about blue shifts near the Cauchy horizons to be true for all space-times.
Best Answer
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.