Interpretation of H as T goes to absolute zero
Short answer: No. There's no reason to expect that this should be the case.



My standard reference for classical thermodynamics is Callen's Thermodynamics and an Introduction to Thermostatistics, 2nd ed. Ch. 11 is relevant here.

Planck's statement of the third law is, according to Callen, "The $T=0$ isotherm coincides with the $S=0$ adiabat." In addition to the theoretical ramifications of having an isotherm and an adiabat coincide, the Planck statement also assigns to this adiabat $S=0$. This is physically grounded in statistical mechanics.

Let $\Delta$ represent the change in a quantity of a reversible isothermal process. Then, at zero temperature:


$S = 0$, by the third law.
$\Delta S = 0,$ again by the third law.
$\Delta G-\Delta H = T\Delta S = 0,$ because $T \to 0$.
$$\frac{\text{d}\Delta H}{\text{d}T}-\frac{\text{d}\Delta G}{\text{d}T} = \lim_{T\to0}\frac{\Delta H-\Delta G}{T} = \lim_{T\to0}\Delta S = 0,$$ once again by the third law. (This shows why the Gibbs free energy and the enthalpy are essentially equivalent thermodynamic potentials at low temperatures!)


But there is no reason to expect $\Delta H = \Delta G = 0$. Regardless of temperature, some energy is still required to effect a change in the system.