After reading this NASA article about the "blackest material", the following stuck out to me.
The tiny gaps between the tubes absorb 99.5 percent of the light that hits them
Is it possible to create a material that absorbs, not just all visible light, but all electromagnetic radiation?
Answer
We can make it absorb a lot of energy but if you read about black-body radiation effect you will notice that as energy is introduced into the object it will similarly radiate a small amount of the energy back, at room temperature appears black, as most of the energy it radiates is infra-red and cannot be perceived by the human eye. At higher temperatures, black bodies glow with increasing intensity and colors that range from dull red to blindingly brilliant blue-white as the temperature increases.
This means that later even if we did have an $100$ percent absorbing materials of all electro-magnetic radiation the object will radiate some energy due to heating or other similar process. That in mind, it will always leak out some energy.To conclude, we could say it will only absorb radiation for short-time before releasing enough photons which could be detected by an photon detector, therefore "breaking" the truly blackest or absorbing material due to this effect.
As a summary as long as the object absorbs some form of energy it can never be completely black.
To learn more pertaining black-body radiation, read this: http://en.wikipedia.org/wiki/Black-body_radiation
Furthermore, an blackhole is very black however it even is not as the power in the Hawking radiation (if it was proven, currently it is hypothesized) from a solar mass $M$ which is equal to $1.98855\pm 0.00025 * 10^{30}$ black hole turns out to be a minuscule 9 × 10−29 watts. It is indeed an extremely good approximation to call such an object 'black'. As $$ P =\hbar c^2/15360\pi G^2M^2 = 9.004 * 10^{-29} $$
That being said, even black-holes one of the strongest objects cannot escape being truly black.
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