Texts on QCD don't divide the generators of $SU(3)$ – and therefore "bicolors of gluons" – into two groups because this separation is completely unphysical and mathematically artificial (basis-dependent).
Moreover, the number of "bicolors of gluons" i.e. generators of $SU(3)$, the gauge group of QCD, isn't nine as you seem to think but only eight. The group $U(3)$ has nine generators but $SU(3)$ is the subgroup of matrices with the unit determinant so one generator is removed. At the level of gluons or Lie algebra generators, the special condition $S$ means that the trace is zero. So the combinations
$$ A(r\bar r) + B (g\bar g) + C(b \bar b)$$
are only allowed "bicolors of gluons" if $A+B+C=0$. Now, in this 8-dimensional space of "bicolors of gluons", there are no directions ("bicolors") that are better than others. For any direction in this space, there exists an $SU(3)$ transformation that transforms this direction into a direction non-orthogonal to any chosen direction you choose. This is true because the 8-dimensional representation is an irreducible one (the adjective "irreducible" means that one shouldn't try to split it to two or several separated collections!). And there doesn't exist any consistent Yang-Mills theory that would only contain the six off-diagonal "bicolors" because the corresponding six generators aren't closed under the commutator.
The actual calculations of the processes with virtual gluons ("forces" between quarks etc.) therefore never divide terms to your two types because this separation is just an artifact of your not having learned group theory. Instead, all the expressions are summing over three colors of quarks, $i=1,2,3$ indices of some kind, and there is never any condition $i\neq j$ in the sums because such a condition would break the $SU(3)$ symmetry.
Now, the $r\bar r,g\bar g,b\bar b$ "bicolors of gluons" (only two combinations of the three are allowed) are actually closer to the photons than the mixed colors. So it's these bicolors that produce an attractive force of a very similar kind as photons – they are generators of the $U(1)^2$ "Cartan subalgebra" of the $SU(3)$ group and each $U(1)$ behaves like electromagnetism. That's why these components of gluons cause attraction between opposite-sign charges and repulsion between the like charges.
The six off-diagonal "bicolors of gluons" (and let me repeat that the actual formulae for the interactions never separate them from the rest – they're included in the same color-agreement-blind sum over color indices) cause neither attraction nor repulsion: they change the colors of the interacting quarks so the color labels of the initial and final states are different. It makes no sense to compare them, with the idea that only momentum changes, because that would be comparing apples and oranges (whether the force looks attractive or not depends on the relative phases of the amplitudes for the different color arrangements of the quarks).
At any rate, particles like protons contain quarks of colors that are "different from each other" so they're closer to the opposite-sign charges and one mostly gets attraction. However, the situation is more complicated than it is for the photons and electromagnetism because of the six off-diagonal components of the gluons; and because gluons are charged themselves so the theory including just them is nonlinear i.e. interacting.
When considering real emission diagrams, one should always check the kinematics first. Your first process $g \rightarrow g+q$ is a perfect example (I use your link here to diagram C without the emitted photon). Gluons are massless, quarks are not and our theory (QCD) is Lorentz invariant. So we choose the rest frame of the initial quark and innediately see, that no gluon can be emitted without violating energy-momentum conservation.
A process like this never happens. That's why there is an additional photon emitted in the diagram linked above and thats why none of your other described processes work. Without further interaction, a gluon (or a photon in QED) cannot simply be emitted by a quark.
Consider QED, e.g. Cerenkov radiation. This phenomenon only seems to be just an electron emitting a photon. However, this can only happen in the vicinity of a nucleus (diagrammatically, this is implemented as an infinitely heavy "background", meaning that the mass $M$ of the nucleus is much larger than the electron mass: $M\gg m_e$.
(This is my old answer conserning loop corrections. I just leave it here, though it does not answer the asked questions.)
By sqaring the Feynman rule (as you put it) you draw a diagram, where you connect one gluon and one quark line. You thus have a loop diagram as a correction to the quark propagator. You can easily implement this, by multiplying with a delta function, that connects two indices, one of each $T$. E.G.
$g_s^2T^a_{ij}T^a_{km}\delta_{jk} = 2 \pi \alpha_s(\delta_{jk}\delta_{im}-\frac{1}{N_c}\delta_{ij}\delta_{km})\delta_{jk} = 2 \pi \alpha_s (N_c \delta_{im}-\frac{1}{N_c}\delta_{im})=4\pi \alpha_s C_F\delta_{im}\ .$
The problem is similar here. You calculate a correction to the gluon propagator with a closed quark loop. Writing the terms in correct order, you see that, since it is a closed fermion line, the firs and last indices are the same, hence you calculate a trace.
Its the same thin again (here you calculate a gluon loop as a correction to the gluon propagator). You have to let go of the notion "squaring Feynman rules". Draw the diagram and add an index to each line. Apply the Feynman rules and you are done (modulo some $f$ acrobatics ;) ).
Best Answer
Experimentally speaking, it is found we detect only $\mathrm{SU}(3)$ colour singlets in nature so the gluons we draw in our feynman diagrams will ultimately hadronise to form such states at time scales longer than the hard scattering with which they were involved. So at the end of the day the resultant states will carry $\mathrm{SU}(3)$ colour quantum numbers with a net red/green/blue configuration. Colour combinations other than this, as carried by the gluons in your picture, are not discernible to detectors so from this viewpoint we expect to consider the two topologically distinct diagrams as drawn.
BTW those two diagrams are also equivalent to