A question about the notation of the Standard model The ${\rm SU(2)_L}$ doublets, for example, the left-handed quark doublets $Q_{iL}\equiv(u_{iL}, d_{iL})^T$ are assigned quantum numbers $(\textbf{3},\textbf{2})_{+1/6}$, which means $Q_{iL}$ are triplets under color ${\rm SU(3)_C}$. Here, '$i$' represents the generation index. On the other hand, the right-handed ${\rm SU(2)_L}$ singlets are denoted as $U_{iR}$, with quantum numbers $(\textbf{3},\textbf{1})_{+2/3}$. But my understanding is that the column 
$$(u^{red}_{i}, u^{green}_{i}, u^{blue}_{i})^T , $$ 
with 3 rows and without the index $L$ is the fundamental representation $\textbf{3}$ of ${\rm SU(3)_C}$. The $\textbf{3}$ representation cannot be the column 
$$(u^{red}_{i}, u^{green}_{i}, u^{blue}_{i},d^{red}_{i}, d^{green}_{i}, d^{blue}_{i})^T , $$ 
with 6 rows.
So my question is the following. Is it not an abuse of notation in which $(\textbf{3},\textbf{2})_{+1/6}$ denotes $Q_{iL}\equiv(u_{iL}, d_{iL})^T$?  Can the QCD Lagrangian be written in terms of $Q_{iL}$ and $U_{iR}$?
 A: 
The 3 representation cannot be the column $\ldots$ with 6 rows.

This is a basic misunderstanding about representation theory. Suppose that a group $G$ has a representation $R$, so vectors $v$ in this representation transform as
$$g(v) = R(g) v.$$
Similarly let a group $G'$ has a representation $R'$, so vectors $v'$ in this representation transform as
$$g'(v') =R'(g') v.$$
Then the group $G \times G'$ has the representation $(R, R')$ where
$$(g, g')(v \otimes v') = R(g) v \otimes R'(g') v'.$$
You can check straightforwardly that this satisfies the properties of a representation. In terms of matrices, the representation matrices of $(R, R')$ are simply the tensor products $R(g) \otimes R'(g')$. The dimension of this representation is $\text{dim}(R) \, \text{dim}(R')$.
This shouldn't be unfamiliar. For example, neutrons and protons are spin $1/2$ particles that transform under both $SU(2)$ rotations and $SU(2)$ isospin. The neutron and proton have two spin states each, so an isospin doublet contains $2 \times 2 = 4$ states. Each state is a tensor product of a spin state (up or down) and an isospin state (proton or neutron). 

Is it not an abuse of notation in which $(\textbf{3},\textbf{2})_{+1/6}$ denotes $Q_{iL}\equiv(u_{iL}, d_{iL})^T$? 

No, that's perfectly fine. $Q_{iL}$ is an $SU(2)_L$ doublet of $SU(3)_C$ triplets.

Can the QCD Lagrangian be written in terms of $Q_{iL}$ and $U_{iR}$?

Of course it can. Just take the Standard Model Lagrangian, already written in these variables, and remove all the terms you don't want. In the context of QCD, writing things in terms of weak isospin multiplets is a bit clunky, but it isn't wrong.
