This is one of those cases where everything depends on conventions,
frames of reference and coordinate systems, but most answers are right
from someone's point of view.

In a local inertial frame of reference, mass and energy are related
via E = m c^2. However, a coordinate system from which gravity looks
like a force is not inertial, and in such a coordinate system c
depends slightly on the potential. This makes it a bit tricky to talk
about what happens to energy and mass as seen from a distance.

In the simple case of a static central mass described in isotropic
coordinates, the potential can be approximated as a single scalar
factor, equal to (1-GM/rc^2) in a weak semi-Newtonian approximation.
This scale factor (let's call it Phi) determines the rate of a
standard clock and the size of a standard ruler at a distance r from
the central mass, so c varies as Phi^2.

When a small object falls towards the central mass, as in the
Newtonian model, its total energy as measured in the isotropic
coordinates remains constant, but potential energy is turned into
kinetic energy. The potential energy varies as Phi, and is equal to m
c^2 where m is the mass of the test object (which remains fixed in its
own frame of reference). Since c varies as Phi^2, this means that
relative to the isotropic coordinate system, the rest mass varies as
Phi^-3, which means it actually INCREASES with decreasing potential,
although the corresponding rest energy decreases.

For most purposes, quantities measured in energy units are more useful
than those measured in mass units; energy is conserved, and is closely
related to frequency which can be compared at a distance.

As an object falls towards the event horizon, then its calculated rest
energy does decrease towards zero. However, the overall total energy
and momentum (taking into account the movement of the black hole
towards the object) are unaffected, so this has no physical effect
that could be observed from a distance.