Luminosity of black hole accretion disc The luminosity of a black hole accretion disc gaining mass at a rate $\mathrm{d}M\over \mathrm{d}t$ can be estimated as

$${1\over 12}{\mathrm{d}M\over \mathrm{d}t}c^2$$

That is a substantial proportion of the rest mass of the in-falling matter. The linked document explains that the factor 1/12 is because at less than 3 times the event horizon radius the matter "spirals in without
radiating more energy" 

What causes matter in the accretion disk not to radiate beyond this limit? It is twice the radius of the photon sphere, is the reason general relativity?
Three times the Schwarzschild radius corresponds to the closest stable circular orbit around a black hole. The general idea is that as matters moves in towards the black hole it gets stuck in an accretion disc where angular momentum has to be moved outwards in order to allow the matter to move inwards. The generic mechanism is some sort of viscosity, which heats the gas and hence you get radiation.

However, once the matter gets inside $3r_s$, that problem disappears. There are no stable orbits, no angular momentum loss or viscosity is needed and the material is able to flow (rapidly) straight into the black hole.

Thus when we observe black hole accretion discs we expect them to be truncated at $3r_s$. 

So I think the argument then is along the lines of - the gravitational potential energy of unit mass falling to $3r_s$ is converted into an orbital kinetic energy of $0.5v^2 = GM/6r_s$ per unit mass and the rest is converted to radiation. Thus
$$L = \left[\frac{GM}{3r_s} - \frac{GM}{6r_s}\right] \frac{dM}{dt}$$
$$ L = \frac{GM}{6r_s} \frac{dM}{dt} = \frac{1}{12}c^2 \frac{dM}{dt} .$$