How is conservation of energy addressed when an object reaches a hair outside of the event horizon, if time speed approaches zero? My concern relates to Einstein's rock-down-a-well thought experiment.  If Einstein drops a rock, it picks up velocity and gains in total energy.  If an assistant at the bottom of the well converts the rock into a photon by $E=mc^2$ it will have more energy than if Einstein had done that atop the well.  When the assistant shines the photon back up to Einstein, he must find that the energy has decreased, or he will be able to violate the conservation of mass and energy by forming a bigger rock.  This is one way to show that time must slow with elevation in order to exactly offset energy gained from gravitational potential energy conversion.
So what happens if the assistant is right outside a black hole?  In standard theory, the rock arrives with a less than infinite KE, yet there can be any ratio of time speed we want right up to infinity.  That is, convert the rock into a photon, and shine the photon back to the origin, and you can red-shift it to as small an energy as you want, and eliminate mass and energy from the universe.
Is this a violation of conservation of mass and energy?
 A: The time dilation factor applied to the emitted photon exactly compensates the factor by which the total rock energy has increased by falling at any $r$ and there is no energy conservation problem.
Let's make it simple and assume the rock is originally at rest at infinity, with energy $E_\infty = mc^2$, and falls radially inwards.
A static observer (at $r> r_s$ of course), will observe the speed of the rock as it passed to be $c\sqrt{r_s/r}$.
The rock energy (in the local inertial frame of the static observer) will be
$$E = \gamma mc^2 = \left(1- \frac{r_s}{r}\right)^{-1/2} mc^2\ .$$
If the energy of the rock is turned into a photon of frequency $\nu_e= E/h$ and emitted radially outward, then the photon will be redshifted. The factor by which it is redshifted (according to an observer at infinity) is given simply by the Schwarzschild metric as
$$\nu_{\rm obs} = \nu_e \left( 1 - \frac{r_s}{r}\right)^{1/2} = \frac{mc^2}{h}\ .$$
Thus the photon has an energy of $E_\infty$ and energy is conserved.
A: 
In standard theory, the rock arrives with a less than infinite KE, yet there can be any ratio of time speed we want right up to infinity

These two things, the amount of KE gained and the amount of time dilation are not independent, as you seem to imply. The further you fall the more KE you gain and the more gravitational time dilation you undergo. The two are tied together. The time dilation is always the precise amount needed to compensate for the KE so that energy is conserved. This can be calculated, and it is not trivial but it is straightforward.
