Certainly vinas is correct. The absorbed energy is converted to heat energy.
The scenario you mention with the LED is very close to the blackbody problem known as the "ultraviolet catastrophe." There is a Wikipedia article about it here. What happens in the situation you described is that the light proof box gets hotter. It will increase in heat until the heat leaving the box due to conduction, convection, and radiative effects balance the energy emitted by the LED. Given the typical power required by an LED, the temperature gain would be minimal.
For all practical purposes, all wavelengths of sunlight will heat an object of any color. The amount of heat produced by light incident upon an object depends on the material properties. Some wavelengths will be reflected, some absorbed, and some transmitted.
As you point out, the associated energy of the photon is important in determining what sort of interaction will take place. A photon of a certain threshhold energy will be able to free a bound electron from an atom causing the photoelectric effect. A photon of lesser energy can still free less strongly bound electrons causing the Compton effect. In this case, the photon continues on with less energy. If the energy is high enough, a photon can cause pair production and create an electron and a positron.
The case you are wondering about, where a photon transfers energy but does not release an electron do occur. Depending on whether the photon interacts with the electrons or the entire atom the process is called Thompson or Raleigh scattering respectivly. In the first case, the electron simply returns to the lower energy level and re-emits a photon of the same energy; the result is effectively an elastic scattering event. In the second case, the atom re-emits a photon of slightly less energy in essentially the original direction. The drop in energy of the photons is due to transfer of momentum because of recoil effects.
You are confusing additive and subtractive colour mixing. If you mix paints together you should get black, not white.
In additive mixing (as used in TVs and monitors), you create light, which is then mixed. When you mix the three primary colours (red, green and blue), you produce white. Other mixes produce other colours, for example red and green combine to produce yellow.
When you use paints, you are using an external light source (the sun or a light bulb) and each paint reflects some of the wavelengths and absorbs others. For example, yellow paint absorbs the blue wavelengths, leaving red and green, which mix to yellow. This is called subtractive mixing, and the primaries are cyan, magenta and yellow; when you mix paints of these colours, the result is black. Adding additional colours to this mix keeps the result black, as there is no more light to reflect. Other colours are made up by mixing the primaries.
With both additive and subtractive mixing, the result of mixing colours depends on the purity of the primaries. No paints are "perfect" cyan, magenta or yellow, and as a result the mix will not be completely black. You may get a dark brown or purple, depending on the paints you use. This is one (of several) reasons why printers use black as well as CMY.
The same goes for monitors: you never get "pure white" - which is typically defined as light with a colour temperature of 5500K, about the same as sunlight. Some monitors can be set for different temperatures. Some are set to 9000K, giving white a bluish cast. Interestingly, the colours that can be displayed on a monitor do not match those of a printer (or paint). A monitor can display colours that a printer cannot print, and vice versa. Every device has its own colour gamut, usually smaller than the eye's gamut, so with any device there are colours we can see but which the device cannot produce.
The reason why all this mixing occurs is because our retina has sensors for red, green and blue, and the brain mixes these inputs to tell us what colour we are seeing. This is why the primaries are RGB, or CMY.
Best Answer
If you are considering a single isolated atom then it's true that the atom has no way of getting rid of the energy from the photon except by emitting another photon. However as soon as the atom is surrounded by other atoms there are various mechanisms for radiationless decay i.e. transferring the energy of the absorbed photon into channels that don't involve reradiating the photon.
In a gas the excited atom or molecule can collide with another atom/molecule and transfer the excitation energy into kinetic energy. This is known as collisional de-excitation (that Wikipedia article is for collisional excitation, but de-excitation is the same process in reverse).
In a solid the energy can be transferred to lattice vibrations, i.e, heat, which is generally known as quenching. In fact in most solids quenching is so efficient that almost no energy is reradiated as photons. Reradiation in fluorescence or phosphorescence is the exception rather than the norm.