When a spacecraft is on a reentry trajectory, its kinetic energy is roughly
$$E_k=0.5(8000{m \over s})^2=32 MJ/kg$$
All of that kinetic energy needs to be converted to heat to slow the spacecraft down. With a well-designed heat shield, such as the Space Shuttle's, about 90% of the heat generated during reentry will end up in the atmosphere (ref), and 10% will be absorbed by the spacecraft. You want the absorbed heat to go into the thermal protection system's (TPS's) mass (as the TPS is supposed to protect the rest of the spacecraft and its payload from heating up). So the thermal protection system must have enough heat capacity to absorb 10% of 32 MJ/kg. This works out to 3.2 MJ of heat capacity per kg of spacecraft.
The Latent Heat of Vaporization of Hydrogen is 0.44936 kJ/mol. Hydrogen is 0.002 kg/mol, so that works out to just
$$0.44936/0.002 = 224 kJ/kg = 0.224 MJ/kg$$
The heat capacity of H2 is ~29 J/mol*K, so if we assume that we raise the temperature of the H2 by, for example, 1000K after we vaporize it, that gives us an additional
$$29/0.002*1000=14.5 MJ/kg$$
Altogether we can store 14.5+0.224=14.724 MJ of heat in each kg hydrogen.
If all of the absorbed heat energy is absorbed by H2 coolant then, if we need to get rid of 3.2 MJ of heat per kg of spacecraft, we will need
$$3.2/14.724=0.217kg$$
of H2 per kg of spacecraft.
Since that extra H2 just added weight to the spacecraft we will actually need roughly
${0.217 \over (1-0.217)} = 0.277$ kg H2 per kg of spacecraft
(Note: You will arrive at different numbers if you make different assumptions. Consider this to be an illustrative calculation.)
I'm assuming here that it would be best to just vent this hydrogen. To burn it in the engines you would need to bring along extra O2 as well, which would add weight as O2 has a much lower heat capacity. If you tried to store it you would need stronger tanks which would also add weight.
Incidentally, you can calculate a rough "equivalent exhaust velocity" for the coolant hydrogen consumed during aerobraking by using the rocket equation.
$${\Delta}V/v_{eq\_exhaust}=ln(m_0/m_f)$$
$$v_{eq\_exhaust}={{\Delta}V \over ln(m_0/m_f)}={8000 \over ln(1.277/1)}=32,718 m/s$$
Now you can use this $v_{eq\_exhaust}$ value to estimate how it scales to lunar velocities. If the craft's return speed from the moon was 11000 m/s then its $m_0/m_f$ would be
$$m_0/m_f=exp(11000/32718)~=1.4$$
So in the moon-return case, the spacecraft would lose 40% of its mass during reentry by venting coolant hydrogen.
Ablative heat shields might do better because their bits can become white hot before they flake off, so they can carry away lots of heat per kg of ablative material. Hydrogen coolant cannot get any hotter than the maximum allowable temperature for the metal heat shield.
