How do we determine the temperature of a Black Hole? Since we cannot see a Black Hole, which I presume, is because it absorbs light, would it not also prevent radiation from escaping, making temperature determination difficult.
Classically, you're exactly right.
Quantum mechanically, something strange happens. It turns out that the way that we define what a particle is and, therefore, what the vacuum is, is dependent on our notion of time--in particular, you are stuck with particles being objects that propagate information forward in time, rather than backward in time.
In general relativity, though, we know that the notion of time on a black hole horizon is inequivalent to that far from the black hole. What this means is that if you take a vacuum solution far from the black hole, and in the distant past of the black hole, and trace it forward in time, you will find that by the time you reach the black hole, it will appear that that state will contain particles! Furthermore, some of these particles will be travelling above escape velocity, and at late times, will be arbitrarily far from the black hole. So, Hawking radiation is inherently a quantum mechanical effect deriving from the fact that the definition of a 'particle' is ambiguous in general relativity.
Furthermore, when you analyze this gas of radiating particles, you will find that it is exactly equivalent to a gas having a temperature that is proportional to the gravitational force felt at the horizon by an observer falling into the black hole.
The temperature is determined using the equivalence principle, and the temperature detected by accelerating detectors.
When a detector is accelerating forever in a certain fixed direction with acceleration $a$, it sees a thermal background temperature equal to $a/2\pi$. This thermal background is associated with the horizon that is formed by the accelerating particle, since a beam of light a certain distance away from the accelerated observer never catches up.
The reason is that an accelerated path is a hyperbola, because the acceleration is the curvature of the particle's path through space-time, and the curvature is constant, and in relativistic geometry, the constant curvature curve is a hyperbola, just as in Euclidean geometry it is a circle (this comes from the minus sign in the relativistic version of the pythagorean theorem).
The hyperbolic time coordinate is the analog of the angle in Euclidean polar coordinates, and if you formulate a quantum field theory for the accelerating coordinates, the imaginary time continuation is periodic with period $2\pi$, which means that the theory is thermal. This is a little technical, but you can just take Unruh's result on faith. For a free field theory interacting with a detector, like photons interacting with an accelerated atom, Unruh's result that the atom behaves as if it were in a thermal environment can be verified by direct calculations of the response of an accelerated detector, not just by fancy methods.
If you are near the horizon of a black hole, you must accelerate to keep from falling in. This means that you must see an Unruh temperature, by the equivalence principle. This Unruh temperature is in the "t" coordinate of the exterior, which turns into the Unruh angle-like coordinate near the horizon. This radiation must be self-consistent as you move away from the black hole, and the self-consistency condition is that the radiation is at a temperature which is smaller by an amount proportional to the rate at which clocks tick. the faster they tick, the colder the temperature.
When you redshift the near-horizon Unruh temperature to infinitely far from the black hole, you find a finite answer. This answer is $8\pi M$, or proportional to the mass of the black hole. This is equivalent to finding the analytic continuation of the black hole solution to imaginary time, and identifying the periodicity of this solution, and the reason is the physical argument above (since Unruh radiation is periodicity in imaginary time).
This argument has in the past been confusing, because most solutions do not have an analytic continuation. How does a physical temperature depend on analytic continuation? It doesn't really. The only thing that is used is the equivalence principle, to link the periodicity in imaginary time near the horizon (where it is just Minkowski space) to the periodicity far away. Mathematically, this is the same as the periodicity in imaginary time of the continued general relativistic solution.