To see where QCD starts to differ from the behavior of EM fields, we might begin by looking at the classical field. A search brings up [question 339978] and [question 360061] but no answer is found there. With the usual [QCD Lagrangian] definition: $$\begin{align} \mathcal L_{\text{QCD}} &= \bar\psi(i{\large \not}D-m)\psi -{ \frac{_1}{^4}} G^a_{\mu\nu}G_a^{\mu\nu} \\[6pt] D_\mu &= \partial_\mu - i g\ T_a A_\mu^a \\[6pt] G^a_{\mu\nu} &= \partial_\mu A_\nu^a - \partial_\nu A_\mu^a + g f_{abc} A_\mu^b A_\nu^c \end{align}$$ we can use Euler-Lagrange, $ {\small \partial_\mu}\frac{{\large \delta} \mathcal L}{{\large \delta}\partial_\mu A_\nu^a} = \frac{{\large \delta} \mathcal L}{{\large \delta} A_\nu^a} $, to find field equations: $$ \partial_\mu G^{\mu\nu}_a = J^\nu_a + g f_{abc} G^{\nu\rho}_b A_\rho^c $$ where $J$ is the source current from the matter field only (the second term shows how the field also acts as its own source). My question is: what goes wrong if I assume that the matter field gives us only static color charge density of one kind, say $a=1$? So the only component of the source current is $J^0_1$. It seems that we then get exactly the Coulomb solution for the color-electric field $G_1^{\,0i}$ with $i=1..3$, and we can even express it simply using an electrostatic potential: $$ V \equiv A_0^1, \quad\text{with}\ \ \nabla^2 V = -J^0_1, \ \ E_i^1 \equiv G_{i0}^1= -\nabla_i V, \quad\text{etc.}$$
As long as there is only a component of $A$ for one value of the $SU(3)$ index, the terms $f_{abc} A_\mu^b A_\nu^c$ will always be $0$ because the structure constants $f_{abc}$ are antisymmetric. So we get ordinary electrostatics back and there is no confinement! (Likewise, if we would make it non-static, we would get electrodynamics back if there are only the components $J^\mu_1$.)
Is this perhaps impossible because the fermion field can never create a charge density $\bar\psi \,T_a\,\psi$ for only one value of $a$? If $\psi$ is in the ${\bf 3}$ representation of $SU(3)$ that seems plausible, since it has only 3 (complex) degrees of freedom and can never create any arbitrary point in the 8-dimensional space of $SU(3)$ generators. But what about the ${\bf 8}$ or the other higher representations? Or is there something else we have to change? Do we have to go to QM? The question then seems to become: what is the simplest (but not simpler) analysis that we can use to see confinement.
