From Craig Hogan's paper 0706.1999:

"[W]ere there no holographic uncertainty, it would be possible 
to find many distinguishable inward trajectories for particles 
[...] that would end up creating exactly indistinguishable black 
hole states. [...] Holographic uncertainty offers a way to 
resolve the contradiction in a way that respects the continuity 
of the trajectories of evaporated particles. The distant 
spacetime does not have so many independent degrees of freedom: 
all those apparently dierent inward traveling particle states 
are not actually independent states. The problem is solved for 
black holes of any size if the holographic limit on the number 
of angular states as a function of distance applies to at spacetime 
generally." 

Does one really need a modification of the usual uncertainty
to avoid such a contradiction?

Suppose you have a shell of particles that will be forming the 
black hole. How many dof can these have? Consider one 
particle in place (0,0,0) and the  black hole would be formed 
in z direction, after the particle had traveled distance D (and 
there met with all the other particles to collapse). Thus, the 
particle needs to be aimed precisely enough to end up inside 
the Schwarzschild horizon. The uncertainty in momentum in the 
transverse directions thus  has to be

\Delta p_x / p_z < R_H/D

where p_x is the momentum in x-direction etc, and a
similar relation for the y-direction. (This isn't exactly
right as the curves of the particles are not tangential
to the Schwarzschild sphere, but actually intersect it at
a shorter distance, but at distances much larger than the
radius this is an arbitrarily good approximation). 

Plugging in the usual Heisenberg uncertainty, one finds 

\Delta x > D/( R_H p_z )

Squaring this, we can devide up the shell into the areas
determined by this uncertainty and find that the number of
patches that fit on the sphere (N_S) is 

N_S < D^2/Delta x^2 < R_H^2 p_z^2

Where p_z should be understood as the radial coordinate now.

Now from Hogan's argument, the number of dof on the shell
should not be larger than that of the black hole formed,
that would be N_BH = R_H^2/l_p^2. Thus one has a problem
if

N_S > N_BH                  (*)

which, putting in the above leads to p_z > m_p (where m_p =
1/ l_p is the Planck mass). 

However, if I have the shell with all these particles, each with
momentum > m_p, then the mass (M) of the black hole I will
be creating must be larger than the number of particles times
at least this energy, thus

M > R_H^2 p_z^3 > R_H^2 m_p^3 = M^2/m_p

or m_p > M, meaning (*) only poses a problem if the mass
of the black hole is below the Planck mass. 

That is as far as the creation of a black hole is concerned.
Looking at the evaporation of a black hole, for one the
question how much information could have been encoded in
the radiation when measured at a very large distance
doesn't change how many indeed were encoded, but besides
this the average energy of the emitted particles is ~
1/R_H which is < m_p again as long as M < m_p. 

Thus it seems the usual uncertainty relation is sufficient to
deal with the problem Hogan is concerned about.

?