Is a 6-quark particle viable? It is my understanding (which may be flawed) that protons and neutrons are stable because the 3 (R, G, and B) quarks form a "white" color singlet.  Wouldn't 6 quarks or even 9 quarks create a white singlet?  What about RGBGR?
 A: The deuteron (or $^{2}$H nucleus) is a color-neutral bound state of three $u$ quarks and three $d$ quarks.  Because six quarks can combine to make a color-neutral state, this six-quark state is possible.  However, the quarks inside are not arranged in anything like a symmetric state. They are tightly bound into two substructures which are already themselves color neutral, one with composition $uud$ and the other $udd$; these are just the proton and neutron constituents of the deuterium nucleus.
There are also nine-quark structures, the nuclei of $^{3}$H and $^{3}$He.  The former is unstable, but its decay lifetime is a matter of years, not the yoctoseconds typical of really unstable strong states.  The latter is stable outright.  As in the deuteron, these states are almost entirely combinations of smaller color-neutral nucleons.
A: In principle, a hadron with any number of quarks can be formed provided that the overall color is neutral.
However, hadrons with more than three quarks (observed in particle accelerators) are unstable$^1$ and decay rapidly.
Tetraquarks and pentaquarks have been observed in high energy collisions, but rapidly decay. A particle of the form $\mid RGBGR\rangle$ does not appear to be viable since the overall color is not neutral.
However, something like $\mid RGBG\bar G\rangle$ where $\bar G$ means anti-green, could be a possible quark "molecule" or pentaquark, since we have $$\mid\underbrace{RGB}_{neutral} \ \underbrace{G\bar G}_{neutral}\rangle$$ or four quarks and one antiquark bound together.
As per the link, a tetraquark, or "meson molecule" (two mesons) the combination $$\mid\underbrace{q\bar q}_{neutral} +\underbrace{Q\bar Q}_{neutral}\rangle$$ is also possible, provided we have color-anti-color, or net color neutral (the bar here means anti-color and not antiparticle).
There has in fact been experimental evidence at CERN for a six-quark state, “dibaryon” or Hexaquark.
It would appear that provided we have enough energy, the synthesis of a fleeting hadron with any number of quarks is possible, once again, only if the net color of the combination is neutral.
As for a "nine-quark state", it could possibly be a "tribaryon" or something of the form $$\mid\underbrace{RGB}_{neutral} \ \underbrace{RGB}_{neutral}\ \underbrace{RGB}_{neutral}\rangle$$ but you would need a lot of energy to synthesize this particle, and I do not think anyone has observed a particle with greater quark number than a dibaryon (six).
$^1$ This is because there are more stable lower mass hadron states that they can decay into.
