So I got into a mini-debate in science class today because I proposed that black holes aren't really black, they only look black because light can't reflect off them. But if you were to take the material that makes up that black hole and decrease it's density so that gravity isn't so strong that light can't reflect, then what color will this new object be? Think of it like this: if you put a red apple in a room that is completely pitch black, the apple will appear black but it's actually red (we just can't observe this because there's no light).
[Physics] What Color Are Black Holes Really? (Yes, a serious question)
black-holesvisible-light
Related Solutions
Under normal circumstances, what you are seeing is the steady state condition where the rate of absorption is equal to the rate at which energy is conducted away or radiated away again, so the material doesn't heat up. As I understand, unless a material fluoresces, the de-excitation happens in the infrared.
Also, you should realize that just because a photon being absorbed means that an electron went from energy $E_1$ to energy $E_{10}$, there is no reason to believe that in de-excitation, the electron will go back from $E_{10}$ to $E_1$ again in one jump. The energy levels of a solid are very complicated (band structure) and has very many finely spaced energy levels that are almost like a continuum. Where there are gaps in the band structure, the material is transparent.
Here is an example of the band structure of a random material (density of states of gold in the middle). I believe colour has more to do with the conduction bands (the upper squiggles) and the relative number of available states.
The conformal limit
For simplicity, consider non-rotating compact objects. A non-rotating object with mass $M$ becomes a black hole when its radius $R$ is $$ R = 2\frac{GM}{c^2} \tag{1} $$ where $G$ is Newton's gravitational constant and $c$ is the speed of light. Equation (1) is the Schwarzschild radius. According to ref 1, in order to avoid becoming a black hole, the radius of a compact object must be $$ R\gtrsim 2.83 \frac{GM}{c^2}. \tag{2} $$ Equation (2) is the conformal limit (ref 4), sometimes also called the causality constraint (but beware that the latter name is also used for something different).$^\dagger$ It comes from the equation of state for ultrarelativistic particles (ref 2), where the pressure $P$ and density $\rho$ are related to each other by $P=\rho c^2/3$. This, in turn, means that the speed of "sound" in a compact object is limited by $v\equiv \sqrt{dP/d\rho}\leq c/\sqrt{3}$, which limits how quickly one part of the object can react to changes in another part, which in turn leads to the bound (2). This bound is consistent with observation (ref 1).
$^\dagger$ The condition $v<c/\sqrt{3}$ is called the "causality constraint" in ref 2 and is called the "conformal limit" in ref 4. Other papers use the name "causality constraint" for the looser condition $v<c$.
This puts a limit on (non-rotating) "almost black holes": the radius must be at least 40% greater than the radius of a black hole. Presumably a similar limit can be derived for the more-realistic case of rotating compact objects, but I'm not familiar with it.
Both the Schwarzschild radius (1) and the conformal limit (2) are indicated near the upper-left corner of this mass-versus-radius figure from ref 3:
The Schwarzschild radius is the boundary of the dark blue region (labeled "GR" for General Relativity"), and the conformal limit (labeled "causality") is the boundary of the upper-left green region. The black curves are various models for neutron stars, and the green curves are models for quark stars.
The Buchdahl bound
Equation (2) comes from considering the equation of state for ultrarelativistic particles. If a realistic equation of state could exceed the conformal limit, then maybe the conformal limit (2) could be beaten. Table 2 in ref 4 suggests that this might be possible. I'm not familiar enough with that work to comment on how realistic that is, but in any case we still have the Buchdahl bound. The Buchdahl bound comes from requiring that the pressure at the center of the object is finite and that the density decreases away from the center (ref 2). The Buchdahl bound is $$ R > \frac{9}{4}\,\frac{GM}{c^2}, \tag{3} $$ which says that the radius of an "almost black hole" must be at least 12% greater than the Schwarzschild radius.
This again assumes a non-rotating object. I don't know what the generalization is for a rotating object.
Bending of light
As explained in ref 5, if light comes close to a certain critical radius of a sufficiently compact object, the gravity can be so strong that the light loops around the object arbitrarily many times before leaving the vicinity, and it can leave in any direction (depending on the precise details of how close to the critical radius). That critical radius is $3 GM/c^2$, 50% larger than the Schwarzschild radius, so an object as compact as (2) or (3) would show this effect. Here's an example from figure 3 in ref 5:
The shaded area is a circle with the Schwarzschild radius (so the compact object will be a bit larger than this), the dashed line shows the critical radius (equations (2) and (3) represent objects smaller than this), and the solid line is the trajectory of the light. The same paper also includes several other figures illustrating various light-bending effects due to such a compact object.
The idea of searching for neutron stars (and other compact objects) using their gravitational-lensing effect has received some attention. Ref 6 is one example.
References:
Lattimer and Prakash, "Neutron Star Observations: Prognosis for Equation of State Constraints", https://arxiv.org/abs/astro-ph/0612440
Eksi, "Neutron stars: compact objects with relativistic gravity", https://arxiv.org/abs/1511.04305
Lattimer, "The Nuclear Equation of State and Neutron Star Masses", https://arxiv.org/abs/1305.3510
Li et al, "Neutron star equation of state: Exemplary modeling and applications", https://www.sciencedirect.com/science/article/pii/S2214404820300355
Kraus (1998), "Light Deflection Near Neutron Stars", https://www.spacetimetravel.org/licht/licht.html (includes a link to download the PDF file)
Dai et al, "Gravitational microlensing by neutron stars and radio pulsars", https://arxiv.org/abs/1502.02776
Best Answer
The material part of a black hole is (classically) compressed into a zero volume area, and almost all of the information of the matter that eventually became the black hole was dissipated away, so the original notion of your question is unanswerable.
There IS another sense in which we can think of your question, though. Black holes are known to shine light through a process known as Hawking radiation, which has a blackbody spectrum. To first approximation, stars are known to also have a blackbody spectrum.
The key point there is that the blackbody spectrum's color is determined by the temperature of the distribution. In the case of stars, this means that the hotter the star, the bluer the color of the star. For black holes, the more massive the star, the redder the blackbody distribution. For black holes that have masses anywhere near that of the sun, the "color" of the star will be very, very far into the radio edge of the light spectrum, and therefore, the black hole will not be visible to the naked eye.