Why can't Pd/C and H2 reduce both the alkene and carbonyl portions of α,β-unsaturated carbonyls?
You hit it right on the nose. The real key piece of information is that given enough time, all the unsaturated bonds will be reduced. This tells you that though the reduction is thermodynamically favorable, it is the difference in the energy barriers ($\ce{\Delta \mathrm{G^{‡}}}$) that prevents the carbonyl reduction from occurring at the same rate as the alkene reduction. This means to produce the alcohol faster, we must manipulate the kinetics of the reaction.

To understand why the carbonyl-reduction transition state is higher in energy, we should consider the differences between the carbonyl and alkene bonds. The bonds are much more polarized, and as such the carbonyl $\pi$ bond is considerably stronger than alkene ($93\ \mathrm{kcal\ mol^{-1}}$ vs. $63\ \mathrm{kcal\ mol^{-1}}$)$^{\mathrm{[1]}}$. This means that at low temperatures, only a small fraction of the molecules have enough energy for the carbonyls to associate onto the $\ce{Pd/C}$ surface. Catalytic hydrogenation of carbonyl compounds (aldehydes, ketones, and especially esters) requires high temperatures and pressures to increase the presence of both the substrate and hydrogen on the catalyst.




$^{\mathrm{[1]}}$ Fox, M. A.; Whitesell, J. K. Organic chemistry; Jones and Bartlett: Sudbury, MA, 1997.