In black body radiation at thermal equilibrium, the absorptivity is equal to emissivity (to conserve energy). Is it fair to say that all of the photons falling on the body are absorbed and emitted? If so shouldn't the emissive radiation depend on the absorbed radiation?

Guide me a bit here...does only rate of emission depend on the absolute temperature and the amount of radiation emitted depend on relative temperature?

In black body radiation at thermal equilibrium, the absorptivity is equal to emissivity (to conserve energy).

Materials have absorptivity and emissivity, radiation doesn't. I would rephrase your statement as:

For any material at thermal equilibrium, the absorptivity is equal to the emissivity

Is it fair to say that all of the photons falling on the body are absorbed and emitted?

All photons falling on a perfect black body are absorbed. The photons emitted from a perfect black body depend only on the temperature of the object (and the surface area).

It does via temperature. If there's a lot of radiation to absorb, the material will increase in temperature until the emission from the body equals that absorbed.

Rate of emission and total radiation emitted (basically the same thing here except for an area scale factor) both depend on the absolute temperature of the material.

There are three separate concepts:

1. a body which absorbs all incident radiation

2. a body which is in a state of internal thermal equilibrium

3. a body which is in a state of internal thermal equilibrium, and also in equilibrium with its surroundings, which have a well-defined temperature

A black body is one which obeys 1. Such a body does not have to be in a state of internal thermal equilibrium, but if it is then it will emit black body radiation at the temperature $$T$$ of the body.

If a black body is in close to a state of internal thermal equilibrium but in surroundings which are either not in equilibrium or are at some other temperature, then the emission from the black body will include a part $$\rho_T$$ which is thermal, and another part which depends on the surroundings. Such a body is not in a true equilibrium, even internally, but it may be that the surroundings are only having a slow effect on it, so its state is moving quasistatically from one equilibrium state to another. If so then one may say it is well-approximated by an internal equilibrium and therefore it is emitting black body radiation.

Finally, if a black body is in equilibrium with surroundings at the same temperature, then the absorption and emission will agree in full and in detail.

Is it fair to say that all of the photons falling on the body are absorbed and emitted? If so shouldn't the emissive radiation depend on the absorbed radiation?

To add to the existing answer, I'll address this point. The incoming and outgoing radiation do not have to match. For instance, the Sun radiates the Earth with energy $$E$$ mainly in visible spectrum, while the Earth re-radiates that same $$E$$ mainly in infrared.

In a perfect reflector (envision a strongly mirrored surface), the incoming and outgoing would match because nothing is absorbed, but this is virtually the opposite of a black body.