Color charge is a general term that describes how a particle transforms under $SU(3)$ transformations, i.e. what is its $SU(3)$ representation.
The terms red, green and blue refer to the fundamental or defining representation of $SU(3)$ which is 3 dimensional. Red, green and blue refer to the three basis vectors in this representation, denoted by $| r \rangle$, $| g \rangle$, $| b \rangle$. This is the representation in which the quarks live so we can assign a red, green or blue color to quarks.
Gluons live in the adjoint representation which is 8 dimensional. We do not introduce a "new" color system for the adjoint representation because there is a wonderful property which allows you to construct the 8 basis vectors of the adjoint representation using the 3 red, green and blue colors of the fundamental representation. We exploit the amazing property (which holds for $SU(N)$ in general),
$$
F \otimes {\bar F} = A \oplus 1 \tag{1}
$$
which states that the tensor product of the fundamental and the anti-fundamental (conjugate fundamental) representation decomposes into the adjoint representation and the trivial representation.
More precisely, the 9 basis vectors on the LHS of (1) are
\begin{align}
| r \rangle | {\bar r} \rangle , | r \rangle | {\bar g} \rangle , | r \rangle | {\bar b} \rangle \\
| g \rangle | {\bar r} \rangle , | g \rangle | {\bar g} \rangle , | g \rangle | {\bar b} \rangle \\
| b \rangle | {\bar r} \rangle , | b \rangle | {\bar g} \rangle , | b \rangle | {\bar b} \rangle \\
\end{align}
The adjoint representation is obtained from this by removing the singlet (trivial) representation from the above which we can do by setting
$$
| r \rangle | {\bar r} \rangle + | g \rangle | {\bar g} \rangle + | b \rangle | {\bar b} \rangle = 0 .
$$
This gives the total of 8 basis states in the adjoint representation. This is the representation in which the gluon lives so there are 8 gluons. However, as I said, we do not introduce 8 new colors to describe these gluons since we can simply combine the 3 fundamental colors red, green and blue.
EDIT - Let me also explain the difference between the $U(1)$ case and the $SU(3)$ case. $U(1)$ is an Abelian group so all of its representations are one-dimensional. To label the representation (aka electromagnetic charge) you therefore need just one number, $n$. Further since $U(1)$ is a compact group, we must also have $n \in {\mathbb Z}$.
On the other hand, $SU(3)$ is a non-abelian group so it has many $k$ dimensional representations for $k > 1$. Given an $k$ dimensional representation, states in that representation are labelled by $k$ numbers, $a_1,\cdots,a_k$.
A quark lives in the 3 dimensional fundamental representation so in general, we need 3 numbers to represent its state. In general a quark state is of the form
$$
| q \rangle = a_1 | r \rangle + a_2 | g \rangle + a_3 | b \rangle
$$
When we say a quark is red, we mean it has labels $(a_1,a_2,a_3) = (1,0,0)$. Of course, a quark in general can be in any superposition of states.
A gluon is in the 8 dimensional adjoint representation so it is labelled in general by 8 numbers
EDIT 2: It looks like I have misunderstood OP's question. It seems they meant to ask the similarity (or difference) between the abelian and the non-abelian transformation laws.
In the Abelian case, the transformation law is
$$
\psi \to e^{ i \theta n } \psi
$$
Here, $\theta$ labels the parameter for the transformation and $n$ labels the representation under which $\psi$ transforms (aka its electric charge).
In the non-Abelian case, the transformation law is
$$
\psi \to e^{ i \theta^a T^a } \psi
$$
Here $\theta^a$ are the parameters for the transformation (analogous to $\theta$ in the $U(1)$ case) and the generators $T^a$ are the generators in the representation under which $\psi$ transforms (analogous to $n$ in the $U(1)$ case).
As an example, if the group is $SU(2)$ then
If $\psi$ transforms in the trivial representation, then $T^a = 0$.
If $\psi$ transforms in the fundamental representation, then $T^a = \frac{1}{2} \sigma^a$ where $\sigma^a$ are the Pauli matrices.
If $\psi$ transforms in the adjoint representation, then
$$
T^1 = \left(
\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & -i \\
0 & i & 0 \\
\end{array}
\right) , \qquad T^2 = \left(
\begin{array}{ccc}
0 & 0 & i \\
0 & 0 & 0 \\
-i & 0 & 0 \\
\end{array}
\right) , \qquad T^3 = \left(
\begin{array}{ccc}
0 & -i & 0 \\
i & 0 & 0 \\
0 & 0 & 0 \\
\end{array}
\right) .
$$
One can similarly write down all the matrices for $SU(3)$ as well but it's longer so I won't bother here (see here).
To make my point once again - the ``charge'' of any particle always corresponds to its representation under the symmetry group. For the $U(1)$ group, representations are labelled by one integer $n$ so we call $n$ the electric charge. In the non-Abelian case, representations are not labelled by just one integer so the labelling is not so simple. In this case, we simply give a name to the representation. Using this language, we would say that the color charge of a quark is "fundamental" and the color charge of a gluon is "adjoint".
Within a representation, there are many states! Again, in $U(1)$ case representations are one-dimensional, so each representation contains just ONE (unique) state. Therefore, apart from the integer $n$ no other information is required to describe this state.
In the non-abelian case, representations are higher dimensional, so in order to describe the state of the particle, one needs more information than just its representation - we need to specify the exact vector. So the color charge of a quark is "fundamental" and its color charge state can be red, green, or blue (or superposition thereof).
