What will happen if a black body at room temperature is dropped into a furnace?? What will it look initially and how in the steady state.? I guess initially it should be black as it is absorbing a lot of heat and heat absorbed is less than heat released. In the steady state should'nt it be not visible as heat absorbed is equal to heat radiated?
Your problem seems more about a wrong understanding the term black body. It doesn't mean "black hole" (in common sense, not considering Hawking radiation), something that only consumes all the light (radiation) that arrives on its surface.
The term "black body" wants to express, that such a body will not reflect the incoming radiation (yes, this sounds pretty much the same as black hole) and whatever you see from the surface of this black body (radiation, light...) has the origin in the body own temperature (here is the difference). No reflections.
So - black body's main feature is that it actually emits spectrum (consult https://en.wikipedia.org/wiki/Planck%27s_law) . At low temperatures it is infrared (and it looks black), at high temperatures, our Sun is also a black body. See the picture from How can it be that the sun emits more than a black body?:
The answer: The black body will will accumulate the heat and it will get to the temperature of the furnace. If the temperature reaches more than approx. 800 Celsius, dark red glow may start to appear. With higher temperature, the color will promote further to red and to yellow. Just imagine piece of metal color when getting heated.
Wood reaches normally 300-600 Celsius, the body would remain black for eye.
Sunspots which look black even though they are at temperature of approximately $3800\,\rm K$ are a good example of a colder body being immersed in a hotter body, the photosphere of the Sun which has a temperature of approximately $5800\,\rm K$.
Eventually your black body will absorb heat from the furnace and when it reaches the temperature of the furnace it will be indistinguishable from the contents of the furnace.