I asked this question a few weeks ago and was dissatisfied with most of the answers I found on the internet, so I eventually managed to procure a copy of Griffiths' excellent text on elementary particles (really, all of his texts are excellent) which includes a section exactly answering my question with what I was looking for. I decided then to answer it myself, in case some other curious person reads this and wants to know.
This is just a very cursory explanation, intended to answer my own question to my own satisfaction.
Griffiths starts by introducing what are basically three copies of EM charge called color charge, and proposes these to be three-element column vectors:
$$c_{red} = \left(\begin{array}{c}1\\0\\0\end{array}\right), c_{blue} = \left(\begin{array}{c}0\\1\\0\end{array}\right), c_{green}=\left(\begin{array}{c}0\\0\\1\end{array}\right).$$
These could in principle could take any vector value whatsoever, except for effects of symmetry in the theory and color confinement.
To figure out how these vector charges interact, we turn to the Gell-Mann $\lambda$-matrices, which are to $SU(3)$ what the Pauli matrices are to $SU(2)$. These are listed by Griffiths, but writing matrices would be a pain; you can look them up on Wikipedia.
Griffiths then takes Feynman scattering amplitudes in lowest order for the chromodynamic interaction, and from these develops potentials for various interactions.
For quark-anti-quark, he has
$$V_{q\bar{q}}(r) = -f\frac{\alpha_s\hbar c}{r}.$$
This is a long-range force in principle, but it is made short-range due to confinement. It takes the same form as the Coulomb potential. The important thing here is the $f$, which Griffiths calls the "color factor". This color factor is like $q_1q_2$ in electrostatics or $\mathbf{p}_1\circ\mathbf{p}_2$ for dipole-dipole forces, and will depend on the color state of the interacting particles in question. It is calculated by
$$f = \frac{1}{4} (c_3^\dagger\lambda^\alpha c_1)(c_2^\dagger\lambda^\alpha c_4),$$
where summation is implied over $\alpha$. Here $c_1$ is charge of incoming quark, $c_3$ charge of outgoing quark, and $c_2,c_4$ charges of incoming and outgoing antiquark.
As an example, Griffiths calculates the interaction between red and anti-blue.
$$c_1=c_3=\left(\begin{array}{c}1\\0\\0\end{array}\right), c_2=c_4=\left(\begin{array}{c}0\\1\\0\end{array}\right).$$
Hence
$$f = \frac{1}{4}\left[(1,0,0)\lambda^\alpha\left(\begin{array}{c}1\\0\\0\end{array}\right) \right]\left[ (0,1,0)\lambda^\alpha \left(\begin{array}{c}0\\1\\0\end{array}\right)\right] = \frac{1}{4}\lambda^\alpha_{11}\lambda^\alpha_{22}.$$
That is, it involves a sum over products of the 1st diagonal element and 2nd diagonal element o each of the Gell-Mann matrices. By looking at their form, the only matrices with both these elements non-zero are the ones labeled by Griffiths $\lambda^3$ and $\lambda^8$. These lead to
$$f = \frac{1}{4}[(1)(-1)+(1/\sqrt{3})(1/\sqrt{3})] = -\frac{1}{6},$$
$$V_{r\bar{b}} = \frac{1}{6}\frac{\alpha_s \hbar c}{r},$$
which is evidently a repulsive force. Griffiths also calculates other interactions. For instance, quark-antiquark singlet interactions, $(1/\sqrt{3})(r\bar{r}+b\bar{b}+g\bar{g})$, which have color factor $f=\frac{4}{3}$ and thus are attractive, explaining confinement of quarks to color-singlet states and the lack of observation for colored states. He also calculates quark-quark interactions, which have a slightly different potential,
$$V_{qq}=f\frac{\alpha_s \hbar c}{r}.$$
As an example, he calculates red-red interaction; it has factor 1/3, hence is repulsive.
There is a lot of this in this very wonderful book, but that's enough to satisfy my curiosity of what color charge is and how it works. Hopefully it is helpful to anyone else. Of course, this was highly simplified for the sake of my own simplified brain and no doubt infuriating to pedants in the field, but if you would like a better explanation and understanding, this was all taken from Chapter 8.4 of Introduction to Elementary Particles by David Griffiths, published by Wiley-VCH, Second Revised Edition -- just to cite sources.
Best Answer
Color-neutral gluons that have the component blue-antiblue do exist, much like red-antired and green-antigreen. However, the sum of these three possible kinds of gluons is unphysical, so there are only two "diagonal" types of gluons. None of these two types of gluons are "genuinely color-blind" or "completely color-neutral".
This is more manifest if you realize that the color dependence of the gluon field may be written as a traceless $3\times 3$ Hermitian matrix. It is traceless because the gauge group is $SU(3)$ rather than $U(3)$ whose dimension is 8 rather than 9. (There are 3 complex entries strictly above the diagonal, which are copied in the complex conjugate way beneath the diagonal, plus 2 or 3 real entries on the diagonal, depending on whether we require the trace to vanish.)
Completely color-neutral gluons, if they were added, would be proportional to the identity matrix and they would couple to all three colors of the quarks equally. In other words, the interactions mediated by such gluons would only depend on the baryon number of the quarks. Experimentally, this interaction doesn't exist. In beyond-the-Standard-Model physics, one may try to extend $SU(3)$ to $U(3)$ in this way (this is very common in braneworld models) but because no new baryon-charge long-distance interaction is seen, the $U(1)$ in the $U(3)$ has to be spontaneously broken at a pretty high energy scale.