Let’s say it loud and clear one more time for the people at the back: “Color charge is not the analogue of electric charge”.
This is one of those misconceptions which is spawned from annoying terminology: “color charge” as referred to in popular science accounts (red, green, blue) doesn’t refer to a charge in the conventional sense. There are a couple of places you may have noticed this: firstly, how come electric charge is observable while color charge isn’t? Secondly, why does electric charge appear in the gauge transformation of a particle coupled to the photon, while no colors appear in that for a particle couples to the gluon? The reason is that these two objects - electric charge and color charge - are completely different. Not many sources mention the difference between the two outright (and popular science accounts love to peddle this incorrect view), presumably leaving it to the reader to discern it for themselves.
$\mathrm U(1)$ representation theory
Let’s recall where electric charge $e$ comes from. The irreducible unitary representations of the group $\mathrm U(1)$ are
All such representations are of course one-dimensional. For a general representation, we have a matrix $K$ rather than a number $k$, which can be diagonalised to yield the charges of its irreducible components as its eigenvalues ${k_1, k_2, k_3, …}$ along the diagonal. Importantly, an elementary particle with charge $+2$ transforms differently than a particle composed of two $+1$’s: the former under $\mathbf 1_{+2}$, the latter under $\mathbf 1_{+1}\oplus\mathbf 1_{+1}$. To recap, to specify an irreducible representation of $\mathrm{U}(1)$, all one needs is an integer $k$. Since the representation is always one-dimensional, there is no explicit “internal” space for an electron: $\mathcal H\otimes\mathbb C^*\cong\mathcal H$.
Skipping ahead a bit, a field in an irreducible $\mathrm U(1)$ representation $q$ transforms as $\psi(x)\to e^{iq\lambda(x)}\psi(x)$ under a gauge transformation, with the gauge field transforming as $A_\mu(x)\to A_\mu(x)-\partial_\mu\lambda(x)$. As usual, we can then define the gauge covariant derivative $D_\mu\equiv\partial_\mu-iq A_\mu$ to construct the gauge-invariant interacting Lagrangian
Generalising to $\mathrm SU(3)$
This is straightforward. Now let’s do the same thing for QCD: we have a quark field $\psi(x)$ (just one flavour for now, the generalisation is obvious). The transformation rules are now
where the $T^a$ form a basis for the $\mathfrak{su}(N)$ representation under consideration. The gauge field (gluon) necessarily transforms in the adjoint (dimension $N^2-1$) representation, but quarks could theoretically be in any representation (they are in the fundamental representation in the Standard Model). How do you characterise a representation in a basis-independent manner? By using the Casimir operators for $\mathrm{SU}(N)$ (most commonly, the quadratic operator $C_2\equiv\mathrm{Tr}(T_aT_a)$ - though this can be degenerate for higher representations, and so necessitates an additional finite number Casimirs to distinguish them). Alternatively, just draw a Young diagram - it’s cooler.
Let’s be standard and take the quarks in the fundamental. This is a 3-dimensional representation (but the Lie algebra has rank 8). That is, the color state (not charge!) of a quark is given by the column vector $(r, g, b)^\top$, i.e. $\vert q\rangle=r\vert r\rangle+g\vert g\rangle+b\vert b\rangle$. This is just a fancy way of saying that the overall Hilbert space of the quark must must involve the tensor product of the color space $\mathbb C^3$.
These are the colors - they label the internal state of the quark in $\mathbb C^3$. Note that any such configurations related by a $\mathrm{SU}(3)$ gauge transformation are identified, so saying that a quark is “red” is completely meaningless. Antiquarks similarly transform in the complex conjugate representation, known as the antifundamental $\mathbf{\bar 3}$, and to emphasise that these are two different representations (even though their vector spaces are canonically isomorphic), we denote its basis by ${\vert \bar r\rangle, \vert \bar g\rangle, \vert \bar b\rangle}$. For both quarks and antiquarks, the $T_a$ above are a set of 8 3x3 matrices, which form a basis for all 3x3 traceless matrices (the fundamental representation of the Lie algebra $\mathfrak{su}(N)$ ). Finally, the total Hilbert space is built as $\mathcal H_\text{tot}\cong\mathcal H\otimes\mathbb C^3_\text{color}$, which gives the fields an internal index $i\in{1,2,3}$.
Next, since the gluons transform in the adjoint, and $\mathbf{3\otimes\bar3=8\oplus1}$, we can use ${\vert r\rangle\vert \bar r\rangle, \vert r\rangle\vert \bar g\rangle, \vert r\rangle\vert \bar b\rangle, \vert g\rangle\vert \bar r\rangle, … }$, with the sole proviso that the singlet $\mathbf 1$ be eliminated by enforcing for instance $\sum_i\vert i\rangle\vert \bar i\rangle\overset!=0$. This time the $T_a$ are 8x8 matrices.
Conclusion
Thus it is the choice of representation, indicated by the choice of $T_a$ (not the dimension of these matrices - distinct representations can have the same dimension!), that is the exact analogue of the electric charge. One ought to say that quarks have “fundamental” charge, anti-quarks have “anti-fundamental” charge, gluons have “adjoint charge”, and baryons have “$\mathbf{10\oplus8\oplus8\oplus1}$” charge. Catchy. Bear in mind that this seemingly “simple” characterisation of QCD charges is deceptive: the way the representations interface during gluon-mediated interactions is far more complicated in non-abelian gauge theory than its abelian counterpart, which gives QCD its distinctive set of exotic phenomena.
The “colors” merely label the internal state of a fundamental quark (or anticolors for an antiquark) - the blame of this misunderstanding lies in the fact that $\mathbf 3$, $\bar{\mathbf3}$ and $\mathbf 8$ are the only representations used in QCD, which means the terms “red, green, blue” get reused everywhere as if they were true charges. If quarks were to transform under $\mathbf{15}$ then none of this confusion would arise.