Is color charge a quantum mechanical observable? If I had 2 pions that were identical, except one was comprised of a red and anti-red, and the other was comprised of a green and anti-green, would I be able to perform an experiment that distinguishes between them?
 A: Color charge in the sense of "being blue, red, green" is not a quantum mechanical observable because the $\mathrm{SU}(3)$ gauge transformations mix the colors. This means it is meaningless to say "We have a blue particle", because we can perform a gauge transformation and then we "have a red particle". Since physical descriptions related by gauge transformations are equivalent, there is no difference between "having a red particle" and "having a blue particle". You cannot, even in principle, determine the "color" of an object in this sense. 
The popular phrasings of "red, blue and green" quarks are actually meaningless. They give a nice heuristic because the "color language" gives a way to draw many intuitive conclusions about otherwise unintuitive group theory, but "red, blue or green" quarks do not exist. Objects in the theory that are related by a gauge transformation are literally the same, there is no difference between a "red quark" and a "green quark" - a quark is a quark is a quark.
What we are able to say (if you manage to deconfine the color-charged stuff, since confinement means that we only see colorless objects) is "I have a color-charged particle", and specify "which kind of color charge" it has, i.e. whether it has merely a color (like quarks), or a color-anticolor (like gluons), or color-anticolor-anticolor (like nothing we know), and so on. (These correspond formally to different $\mathrm{SU}(3)$ representations) This - "color/color-anticolor/color-anticolor-anticolor/..." - is the proper generalization of the $\mathrm{U}(1)$ electric charge to non-Abelian gauge theories.
A: Usually, the charge we refer to in QFT means the Noether charge of some global (i.e physical) symmetry. For example, the Noether charge associated with a global $U(1)$ transformation in QED is called electric charge. One must be careful with this $U(1)$ transformation because lots of people confused it with the $U(1)$-gauge invariance in QED, i.e the redundancy under a local $U(1)$-transformation.
Noether's theorems only applies for global symmetries. You can try applying it to a gauge (i.e local) transformation and find a conserved quantity, but it is not an physical observable because it is not gauge invariant.
More specifically, the conserved quantity you have for a $U(1)$-gauge transformation $A^{\mu}\rightarrow A^{\mu}+\partial^{\mu}\Lambda$ is $$J^\mu=\frac{\partial\mathcal L}{\partial(\partial_\mu A_\nu)}\delta A_\nu=-\frac12 F^{\mu\nu}\partial_\nu\Lambda,$$
which is indeed conserved, but is not gauge invariant.
The same thing goes to $SU(3)$-gauge theory. The colors of quarks are conserved, but are not gauge invariant. In QED, one can essentially view the global U(1) symmetry of a charged fermion as the "global part" of the $U(1)$-gauge invariance. However, in the non-Abelian case, one can easily see that the "global part" of $SU(3)$ is its center, which is a discrete subgroup. In other words, one should not expect a conserved current associated with it.
Instead, given the Lagrangian of a $SU(3)$-gauge theory, $$\mathcal{L}=-\frac{1}{4}\mathrm{Tr}(F_{\mu\nu}F^{\mu\nu})+\bar{\Psi}(iD\!\!\!\!/-m)\Psi,$$
one finds a current $$J^{\mu}=\sum_{a=1}^{N}\left(\bar{\Psi}\gamma^{\mu}T_{a}\Psi\right)T_{a}$$
which is covariantly conserved, i.e $D_{\mu}J^{\mu}=0$, following its equations of motion. Notice that this current is not locally conserved because of the covariant derivative. In other words, it is not a Noether current.
On the other hand, in QCD there are still global symmetries, i.e. Baryon number conservation (i.e global $U(1)$ symmetry) and flavour number conservation etc.
Since @octonian mentioned this, it should be emphasized that here the current $$J^{\mu}=\sum_{a=1}^{N}\left(\bar{\Psi}\gamma^{\mu}T_{a}\Psi\right)T_{a}$$
for non-Abelian gauge theory is not gauge invariant. First of all, the current comes from equations of motion
$$D_{\mu}F^{\mu\nu}=J^{\nu} \tag{1}$$
$$(iD\!\!\!\!/-m)\Psi=0 \tag{2}$$
where equation (1) is the Yang-Mills equation, which is the non-Abelian version of the inhomogeneous pair of the Maxwell equations. Under a $SU(3)$-gauge transformation, one has $$F^{\prime}=UFU^{-1},\quad D^{\prime}=UDU^{-1},\quad\mathrm{and}\quad J^{\prime}=UJU^{-1},$$
which implies that the equations of motion is invariant under the gauge transformation, i.e $$D^{\prime}\star F^{\prime}=J^{\prime}.$$
The covariant-conservation of $J$ can be easily checked:
\begin{align}
D\star J&=DD\star F \\
&=F\wedge\star F-\star F\wedge F \\
&=F^{a}\wedge\star F^{b}[T_{a},T_{b}] \\
&=F^{a}\wedge\star F^{b}f_{ab}^{\,\,\,\,c}T_{c} \\
&=\left\langle F^{a},F^{b}\right\rangle f_{ab}^{\,\,\,\,c}T_{c} \\
&=0,
\end{align}
where in the last line the fact that $f_{ab}^{\,\,\,\,c}$ is anti-symmetric wrt $a$ and $b$ has been used.

To avoid any further misunderstandings, please notice that here the curvautre 2-form is $F=dA+A\wedge A$. Since it was mentioned by @octonion in the comment section, it should be emphasized that $F=0$ does not imply flat connection $A=0$！This is true even in Abelian gauge theory. This is easy t understand if one calculates the Christoffel symbols in spherical coordinates of Minkowski spacetime. That the current $J$ is covariantly conserved is a special property for non-Abelian gauge theory. In contrast, in Abelian gauge theory the non-homogeneous pair of the Maxwell equations read $$\star d\star F=j,$$
and the current $j$ is always locally conserved, i.e $d\star j=0$, regardless of whether the curvature $F$ vanishes or not.
