Is there such a thing as red - anti red gluon? Why such gluon posssibilities appear in equations?
Suppose you have a quark of color state $\lvert q\rangle$. Or equivalently, you could write the color state as a 3-component vector, but for now I'm abbreviating it with a smaller symbol. Anyway, if that quark interacts with a gluon whose color matrix is $T_g$, the outgoing quark has a color state of $T_g\lvert q\rangle$.
A red-antired gluon would be represented by the following color matrix, in the RGB basis:
$$T_{r\bar{r}} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$
If you have a red quark, with color state
$$\lvert r\rangle = \begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}$$
and it interacts with that gluon, the resulting quark will have a color state of
$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \lvert r\rangle$$
so, still red. However, if you naively apply this argument to the red-antired gluon and a green (or blue) quark, you get zero. In practice that means red-antired gluons don't interact with green or blue quarks at all. (The quantum mechanical amplitude for the interaction to happen is related to $\langle q_\text{final}\rvert T_g\lvert q_\text{init}\rangle$.)
You can then extrapolate this to what would happen if you let a red-antired gluon interact with a quark in the state $\frac{1}{\sqrt{2}}(\lvert r\rangle + \lvert g\rangle)$:
$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \propto \lvert r\rangle$$
So the gluon actually changes the color state of some quarks. A white gluon wouldn't do that. White gluons don't change color states; that's what it means to be white.
Now, a true red-antired gluon doesn't exist because the corresponding matrix has a nonzero trace. In a sense, the hypothetical red-antired gluon is part white, but the white part doesn't exist. However you can run through the preceding arguments with a gluon whose color matrix is proportional to
$$\begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1\end{pmatrix}$$
which is physically valid, and the conclusions are all the same. This shows you why this gluon, despite containing colors and anticolors in equal combinations, is not white.
Note that the technical term for what you think of as "white" is "$SU(3)$ singlet".
I'm being a little loose with the notation in this post, but hopefully it makes sense.