There is direct experimental evidence for quarks having 3 colors. Electron-positron collisions produce hadrons at three times the rate that one would expect for colorless quarks.
Particle Data Group - 2010 Reviews, Tables, Plots > Kinematics, Cross-Section Formulae, and Plots > Plots of cross sections and related quantities (rev.). With a center-of-mass energy more than about 1.5 to 2 GeV, one gets a
very good fit with QCD calculations. The lowest-order effect is not QCD, of course, but electromagnetic: virtual photon -> quark + antiquark. The factor of 3 comes from
virtual photon -> red quark + cyan antiquark
virtual photon -> green quark + magenta antiquark
virtual photon -> blue quark + yellow antiquark
Furthermore, electron-positron collisions produce quark-antiquark pairs with the angular distribution expected of quarks being spin-1/2 particles.
http://www.sprace.org.br/slc/file.php/6/HEP-I-II/HEP_6_1.pdf
From the hypothesis that mesons and baryons are gauge-singlet states, one can determine what gauge groups that they may follow. Mesons are quark-antiquark states, and are thus trivial, while baryons are 3-quark states, and are the interesting case. So we must find which gauge group allows 3 quarks to be in a singlet state.
We do this by finding conserved quantities or congruence classes for the simple Lie algebras. Their irreducible representations or irreps are denoted by highest-weight vectors; these are vectors of nonnegative integers. The conserved quantities are calculated from highest-weight vector w by
Q(irrep with w) = (c.w) mod c0
where c0 and the components of c are all nonnegative integers.
All weights in an irrep have a Q value equal to that of the irrep as a whole, and
Q(product of irreps with HWV's w1 and w2) = Q(irrep with w1) + Q(irrep with w2)
Going over the simple Lie algebras:
A(n) - SU(n+1) - c0 = n+1
B(n) - SO(2n+1) - c0 = 2
C(n) - Sp(2n) - c0 = 2
D(n) - SO(2n) - even n: two with c0 = 2, odd n: one with c0 = 4
G2 - none
F4 - none
E6 - c0 = 3
E7 - c0 = 2
E8 - none
For SU(2)/SO(3), this is evident in boson vs. fermion parity.
So to get a c0 that can divide 3, one can only have SU(3n) or E6. Since quarks have 3 color states, that gives us SU(3). The quarks are in the 3 irrep, or {1,0}, the antiquarks in the 3* irrep or {0,1}, and the gluon in the adjoint, 8 irrep, or {1,1}. The conserved quantity here is "triality": Q(w) = (w1 + 2*w2) mod 3. Q(quarks) = 1, Q(antiquarks) = 2, Q(gluon) = 0, Q(mesons) = 0, Q(baryons) = 0.
Furthermore, three (anti)quarks can combine to make a colorless, {0,0} singlet. This combination is antisymmetric, which resolves a puzzle with the quark model. In the baryons, spin and flavor must be combined in symmetric fashion. This is evident from the Delta++ (uuu), the Delta- (ddd) and the Omega- (sss), all spin-3/2 (spins parallel). This violates Fermi-Dirac statistics, expected from quarks having spin 1/2, but being antisymmetric in color yields an overall antisymmetry, thus preserving F-D stats.
The gauge particle of QCD has been named the gluon, and it's also been seen indirectly in electron-positron collision. An energetic quark-antiquark pair makes a two jets of hadrons, one for the quark and one for the antiquark. One of the two can radiate a gluon, and it will make an additional jet. The angular distribution of the resulting gluon can be used to find its spin, and as expected for a gauge particle, its spin is 1.
