It is not completely clear what OP is looking for, but here are some hopefully helpful comments:

1. Classically (meaning when Planck constant $\hbar\to 0$), two fields $A$ and $B$ are super-commutative $$AB~=~(-1)^{|A||B|}BA,$$ where $|A|$ and $|B|$ denote the corresponding Grassmann-parity. In other words, the classical super-commutator $$[A,B]~\equiv~AB-(-1)^{|A||B|}BA~=~0$$ vanishes.

2. The super-commutator in quantum theory is typically a quantum deformation of the classical super-commutator.

3. Note that ghost fields can both be Grassmann-even and Grassmann-odd, depending on the theory.

4. In principle, one may consider superalgebras with several independent $\mathbb{Z}_2$- or $\mathbb{Z}$-gradings $$|\cdot|_1, \quad \ldots, \quad |\cdot|_n. $$ The super-commutator in such a superalgebra is then defined as $$[A,B]~\equiv~AB-(-1)^{\sum_{i=1}^n|A|_i|B|_i}BA.$$ (For instance, one could consider exterior form degree and usual Grassmann-grading as two independent gradings.)

5. A theory may allow for different conventions. The main point is that one should be consistent.

It is not completely clear what OP is looking for, but here are some hopefully helpful comments:

1. Classically (meaning when Planck constant $\hbar\to 0$), two fields $A$ and $B$ are super-commutative $$AB~=~(-1)^{|A||B|}BA,$$ where $|A|$ and $|B|$ denote the corresponding Grassmann-parity. In other words, the classical super-commutator $$[A,B]~\equiv~AB-(-1)^{|A||B|}BA~=~0$$ vanishes.

2. The super-commutator in quantum theory is typically a quantum deformation of the classical super-commutator.

3. Note that ghost fields can both be Grassmann-even and Grassmann-odd, depending on the theory.

4. In principle, one may consider superalgebras with several independent $\mathbb{Z}_2$- or $\mathbb{Z}$-gradings $$|\cdot|_1, \quad \ldots, \quad |\cdot|_n. $$ The super-commutator in such a superalgebra is then defined as $$[A,B]~\equiv~AB-(-1)^{\sum_{i=1}^n|A|_i|B|_i}BA.$$ (For instance, one could consider exterior form degree and usual Grassmann-grading as two independent gradings.)

5. A theory may allow for different conventions. The main point is that one should be consistent.

6. Specifically, concerning fermion matter fields $\psi$, Faddeev-Popov ghost field $c$ and antighost fields $\bar{c}$ in Yang-Mills theory, it is possible to consistently set up the BRST formulation using only one type of Grassmann-grading, in which $\psi$, $c$ and $\bar{c}$ are all Grassmann-odd, and pairwise anti-commuting.

1. Classically (meaning when Planck constant $\hbar\to 0$), two fields $A$ and $B$ are super-commutative $$AB~=~(-1)^{|A||B|}BA,$$ where $|A|$ and $|B|$ denote the corresponding Grassmann-parity. In other words, the classical super-commutator $$[A,B]~\equiv~AB-(-1)^{|A||B|}BA~=~0$$ vanishes.

2. The super-commutator in quantum theory is typically a quantum deformation of the classical super-commutator.

3. Note that ghost fields can both be Grassmann-even and Grassmann-odd, depending on the theory.

It is not completely clear what OP is looking for, but here are some hopefully helpful comments:

1. Classically (meaning when Planck constant $\hbar\to 0$), two fields $A$ and $B$ are super-commutative $$AB~=~(-1)^{|A||B|}BA,$$ where $|A|$ and $|B|$ denote the corresponding Grassmann-parity. In other words, the classical super-commutator $$[A,B]~\equiv~AB-(-1)^{|A||B|}BA~=~0$$ vanishes.

2. The super-commutator in quantum theory is typically a quantum deformation of the classical super-commutator.

3. Note that ghost fields can both be Grassmann-even and Grassmann-odd, depending on the theory.

4. In principle, one may consider superalgebras with several independent $\mathbb{Z}_2$- or $\mathbb{Z}$-gradings $$|\cdot|_1, \quad \ldots, \quad |\cdot|_n. $$ The super-commutator in such a superalgebra is then defined as $$[A,B]~\equiv~AB-(-1)^{\sum_{i=1}^n|A|_i|B|_i}BA.$$ (For instance, one could consider exterior form degree and usual Grassmann-grading as two independent gradings.)

5. A theory may allow for different conventions. The main point is that one should be consistent.