standard model - Group theoretical reason that Gluons carry color-charge and anti-colorcharge

I was wondering how it is possible to see from the $SU(3)$ Gauge Theory alone that Gluons carry two charges colors: $g\overline{b}$ etc.

Some background:

The W-Bosons (pre-symmetry breaking) form an $SU(2)$ triplet and carry the corrsponding weak Isospin $1,0-1$. After SSB/Higgs the charged $W^\pm$-Bosons can be identified with complex linear combinations of the $W^{1,2}$, bosons, and therefore the corresponding term in the Lagrangian is $U(1)$ invariant, i.e. the $W^\pm$ carry electric charge, too.

For a local $SU(3)$ gauge theory 8 gauge fields, the gluon fields are needed. Exactly as it was the case for $SU(2)$, one for each generator $\lambda_a$ and one introduces consequently "matrix gauge fields"

$$ A^\mu = A^\mu_a \lambda_a$$

which can be seen as elements of the corresponding Lie algebra, because the $\lambda_a$ form a basis and the expression above can be seen as a expansion of $A^\mu$ in terms of this basis.

The transformation behaviour is the same for all $SU(N)$ theories

$$ A^\mu \rightarrow U A^\mu U^\dagger + \frac{i}{g} (\partial_\mu U) U^{-1} $$

As usual the fermions transform according to the fundamental representation, i.e. for $SU(3)$ are arranged in triplets. Each row representes a different color as explained in the answer here (What IS Color Charge? which recites from Griffith)

Therefore a red fermion, for example is

$$ c_{red} = \left(\begin{array}{c} f \\0\\0\end{array}\right) $$

where $f$ is the usual dirac spinor. An anti-red fermion would be

$$ c_{red} = \left(\begin{array}{c} \bar f 0 0\end{array}\right) $$

The red fermion transforms according to the fundamental rep $F$, the anti-red fermion according to the conjugated rep $F^\star$. Which is a difference to $SU(2)$, because $SU(2)$ has only real representations and therefore the normal and anti rep are equivalent (why is it enough, that they are equivalent? The conjugate rep for $SU(2)$ is different but considered equivalent because $r = U \bar r U^{-1}$, for some unitary matrix $U$. Any thoughts on this would be great, too), i.e. there is no anti-isospin. I guess this is the reason the $W$ do not carry anti-charge, simply because there isn't anti charge for $SU(2)$.

Now where is the point that we can see that the gluons carry anti-colorcharge and colorcharge? Is it because the matrix gluon fields defined above are part of the Lie algebra and transform therefore according to the adjoint rep of the group $A \rightarrow g A g^\dagger$, which could be seen as transforming according to the rep and anti-rep at the same time (or could be seen as completely non-sense idea from me ;) ) ?

Why does the gluon octed does not get charge assigned like the $SU(2)$ triplet, which would mean the gluons carry different values of one strong charge ? (Analogous to $1,0-1$ for weak isospin of the $W$ triplet.)

Any thoughts or ideas would be awesome!