I was wondering if the temperature of an object affects the amount of radiation it absorbs.
For example, if I have a box that is hotter, will it absorb more energy as compared to the same cooler box?
The rules of Black Body radiation say: no. Assuming the composition of the box doesn't change, its absorptivity is the same regardless of temperature. What does change is the amount of energy it'll radiate, which is a direct function of temperature (black body again).
This often confuses folks, as the spectral absorption curve (i.e. percent of photons at a given wavelength) has the same shape as the spectral emission curve. However, the actual absorption depends on the incident radiation, while the actual emission depends on the temperature of the object. For example, a cold black object will absorb all the visible radiation but emit primarily in the long-infrared.
edit: to respond to the comment, here's a quote from wiki. "A gray body is one where α, ρ and τ are uniform for all wavelengths. This term also is used to mean a body for which α is temperature and wavelength independent." A gray body is not 100% emissive, and a 'color body' will have a nonuniform emission curve, but in general these characteristics are most definitely not temperature dependent.
Real bodies absorb radiation of given wavelength with intensity proportional to their emissivity $\epsilon(\lambda, T)$, which in general depends on the temperature and possibly also on other state variables. Whether the absorption increases or decreases with temperature depends on the kind of body and range of temperatures, so it cannot be stated generally.