Can the $SU(3)$ gauge field be put in geometric algebra terms? According to this article on the spacetime algebra, we know the Dirac spinor can be thought of as an even element of the Clifford algebra over spacetime, which in turn can be thought of as a general transformation on bivectors, since it consists of a dilation (scalar), Lorentz transformation ("rotor"), and duality rotation (complex phase).
Thus, the presence of the $U(1)$ gauge field apparently speaks to the duality rotation specifically.
Also, we know that $SU(2)$ is isomorphic to the 3D rotors, hence spatial rotations.  The double cover is nothing to worry about — it's just because a rotor gets both pre- and post-multiplied (just like the wavefunction itself in many cases, so that matches) in order to effect a rotation, so the angle of the rotor gets doubled into the rotation angle.
So what about $SU(3)$?  It is 8-dimensional, just like the even sub-algebra itself, which might make me think it could represent an entire bivector transformation, but that doesn't make sense because it is normalized whereas the dilation part needs to range to infinity.  That is, the even sub-algebra really has 7 rotation-type DOF and one magnitude.
I know that phrasing these gauge operations in GA language might end up going nowhere, but I found the interpretations of the first two symmetries to be interesting, so I was wondering if something in that spirit could be found for $SU(3)$ as well.
 A: The 6D geometric algebra bivectors span
$$spin(6)$$
that is isomorphic to Pati-Salam's
$$SU(4)$$ which in turn contains
$$SU(3)_{color} * U(1)_{B-L}$$
A: Spacetime symmetry and internal symmetry are two separate issue.
When we talk about spacetime symmetry, we say that a field is a scalar field (trivial representation), a spinor field ($(0,1/2)$ or $(1/2,0)$ representation for weyl spinor and $(0,1/2)\oplus (1/2,0)$ for Dirac spinor), a vector field  ((1/2,1/2) representation). The notation $(S_1,S_2)$ label two spin-S representation of SU(2) because the complexification of Lie algebra of Lorentz group $SO(3,1)$ is isomorphic to $SU(2)\times SU(2)$. When we make Lorentz transformation $\Lambda$, the field changes as $\Lambda\psi_i(x)=U(\Lambda)_{ij}\psi_j(\Lambda^{-1}x)$ and $U(\Lambda)_{ij}$ is the representation matrix.
However, the internal symmetry is another thing. It just mean that at every point we make a transformation on the internal degree of freedom $\psi_a(x)\to (U_g)_{ba}\psi_b(x)$. A field could carry several indice, for example for a Dirac fermion couple with a $SU(2)$ gauge field (in  reality this is weak interaction) we write it as $\psi_{ia}$  and when we consider action of Lorentz group the first index changes and $SU(2)$ symmetry the second index changes. The $SU(2)$ gauge group do nothing with the spin or physical rotation of this fermion, and the total symmetry group of it is $SO(3,1)\times SU(2)$
