Is emitted light fundamentally different from reflected light in photography?

With emitted light you work in the additive colour domain, and reflected light is subtractive colours. See the differences in RGB (screens) and CMYK (printers), e.g. to see yellow you can emit R+G, or subtract B.

Emitting a colour is easier to control than subtracting, because reflected light depends on the product of the (emitting) light source and the material BRDF.

BRDF in General

BRDF applied to leaves, where it is interesting how visible light and NIR light acts differently, which is useful in agricultural engineering. I used this theory myself to successfully make an algorithm that can detect scab on apple leaves.

The strength is controllable in both cases, so you can't see that emitted light is stronger. The perceived strength is a function of the wattage you burn off from the light source, distance and spread, and the integral in your viewer sensitivity spectrum of the light spectrum. This is why LED light seems stronger for the same amount of wattage as a halogen light. The halogen does emit more light, but a lot of it is outside the visible spectrum, and is thus not integrated.

If your reflective surface is very diffuse, you get more spread , and are less likely to see a imprint of the light source. this is the reason diffuse lighting is easier to achieve by emitting light into a dome with a white diffuse coating. Perfect white is hard to achieve so this does have some loss, and the distance form the emitting light to the subject will get longer, too. so to get the same strength as you'd get with emitting light directly, you increase the power. You can also reflect the light on a mirror surface and achieve a longer distance, without polarisation or diffusing it. This is often used to increase the FOV of the light source and the physical dimension does not allow it.

Polarisation you can control in both cases. You can get a polariser sheet to place in front of your light. Or you can carefully choose your reflective surface to give the polarisation you want.