What happens when a blackbody absorbs light that does not have a blackbody spectrum?

An object absorbs some fraction of the incoming radiation. If the object is indeed an ideal black body, all the power is absorbed, and none is reflected.

The body will heat up, and emit a black body spectrum corresponding to its temperature. The emitted power is given by $P=A\sigma T^4$ (this is known as the Stefan–Boltzmann law).

When the absorbed power is equal to the emitted power, the temperature of the blackbody stops changing. By equating the two expressions, you can solve for temperature. So, if the incoming power is $P_\text{in}$, then the steady-state temperature of the blackbody is

$$T_\text{steady state} = \left( \frac{P_\text{in}}{A \sigma} \right)^{1/4} \, .$$

Note that a real body (not an ideal black body) has a finite, wavelength-dependent reflectivity. In other words, real objects don't absorb all of the incoming radiation.

I think you're confusing it with the equilibrium state where the incident radiation is exactly the same as the emitted radiation. An ideal blackbody absorbs all radiation and re-emits it with the spectrum given by Planck's Law. You consider this equilibrium state only to calculate the conversion factor between the energy density and the radiated power (Wikipedia and related question).

In your case, the spectrum is irrelevant. The temperature of the blackbody will reach an equilibrium at $T = (I/\sigma)^{1/4}$, where $I$ is the total power absorbed per unit surface area of the blackbody (for all frequencies) and $\sigma$ is the Stefan–Boltzmann constant. This is known as the Stefan–Boltzmann law.