Yes, the color charge of a quark can be represnted by a vector in $\mathbb{C}^3.$ What's more, if you were to "hold" the quark (which you cannot actually do, but let's put that aside) and move it around a background gluon field, that $\mathbb{C}^3$ vector would be be "rotated" by an $SU(3)$ matrix. This is a lot like how if you parallel transport a tangent vector in Riemannian geometry is ends up getting rotated by an orthogonal matrix. The general principle is the same. More mathematically, if you have a spacetime manifold $M = \mathbb{R}^4$, the gluon field is a connection on the bundle $SU(3) \times M$. (Well, this is all in classical physics. Quantum mechanically, the gluon field is a wavefunctional of connections, with additional subtleties involved.) Now, if there is some "field strength," i.e. you might say there are 'gluons' present, then when you drag your quark in a closed loop it'll come back rotated by an overall $SU(3)$ because there is some curvature in the bundle. Now, a gauge transformation applies a spacetime dependent function $g : M \to SU(3)$ that changes one gluon field to a physically equivalent one. Importantly, the holonomy of dragging the quark in a closed loop will remain unaffected by this gauge transformation, giving us a gauge invariant set objects to characterize the gluon field state with. Note that, while the quark has a $\mathbb{C}^3$ vector, the gluons are best understood as enacting group actions on the quark. This means that their states live in the Lie algebra of $\mathfrak{su}(3)$. Hence the labelling of gluons as, say, "red-blue," which rotates infinitesimally between the red and blue colored quark.