This is a pretty late reply and I hope you've had other answers to a very interesting question.
The first part of my answer is to ask you, if you made gold into a gas, how would you look for its color? What I mean is that gold boils at a very high temperature, ~2805 C. If you heated gold to 2000 C, the pressure of gold atoms (gold gas is monatomic) would be 0.01 atmosphere. That is enough to see a color, but you know that when a solid is that hot, it glows pretty brightly. You see orange or yellow light at that temperature no matter what color material you have heated. This color is due to temperature alone and is called "blackbody radiation". So this means that if hot gold gas is emitting light at all colors (but mostly yellow-orange) because it is hot, it will be hard to pick out its specific color by looking at its light emission.
The second part of my answer is that you can find out the color of hot gold gas if you shine light at specific wavelengths through the hot gas and see which wavelengths get absorbed. (They get absorbed when they have the right energy to move electrons from one orbital to another.) If you Google for "NIST Atomic Spectra Database", click "Levels" and put in "Au I" (neutral gold atoms, Au II is for Au+ ions), you get a lot of spectral lines measured in cm-1. The ones you can see by eye are between violet = 25,000 cm-1 = 400 nm (just do 1/x on your calculator) and red = 14,000 cm-1 = 700 nm. You can see the line at 21,435 cm-1 = 466 nm = blue-violet, near the top. That wavelength of light has the right energy to move an electron from its lowest state, where there are 10 electrons in 5d orbitals and one in a 6s orbital (called 5d10 6s1) to a state with 5d9 6s2. If you were to shine visible white light through gold gas or vapor, everything except that blue-violet color would go through. The color that would pass through gold vapor would be reddish-yellow, much like solid gold.