Why cross section of $e^+ e^- \to \bar{q} q$ is 3 times larger than $e^+ e^- \to \mu^+ \mu^-$? I know the usual answer: quarks carry color charge (let us denote them r,g,b), antiquarks anti color charge, and since the initial state ($e^+ e^-$) does not carry any color charge (and color is always conserved), the only possibility is to pick-up $r\bar{r}$, $g \bar{g}$, or $b \bar{b}$, thus 3 possibilities more than for the production of di-muons. 
Now, I'm not totally satisfied because: since my initial state was colorless, my final state must be a color singlet: $\psi_s = \frac{1}{\sqrt{3}}(r\bar{r} + g \bar{g} + b \bar{b})$. Saying that $r \bar{r}$ is a color singlet (as suggested in my quick un-satifying answer) is not true: $r \bar{r}$ appears in the linear superposition of the color singlet, yes, but also in some linear superpositions of the color octet (since with group theory, $3 \otimes \bar{3} = 8 \oplus 1$). 
So how can I get this famous factor 3 between the two cross-sections? 
 A: The point is that when doing calculations in QCD we only need to calculate the hard process (producing e.g. the $ r \bar{r} $) since non-perturbative dynamics will, with 100% probability, produce color singlets out of the final states. It might provide move intuition to write the relevant Feynman diagrams (assuming e.g. a $Z$ mediator) that produce the full color singlet, $ \frac{1}{ \sqrt{3} } \left( \left|  r \bar{r} \right\rangle  + \left|  g \bar{g} \right\rangle  + \left|  b \bar{b} \right\rangle  \right) $,

The cross sections are only affected by the connected parts, since in non-perturbative QCD, it doesn't "cost" anything to pop $q \bar{q}$ pairs from the vacuum. Here we have 3 diagrams, and hence a factor of $3$.
One might be worried however, that the cross-sections are lowered since the hard $q\bar{q}$ pair might hadronize into two colorless pieces (e.g. two mesons). This is also occurs and all these possibilities need to be included to get the proper "factor of 3 boost". 
A: I post a new version of my answer since I'm now convinced that my previous answer was clearly wrong. Looking at papers dealing with J/Psi ($c\bar{c}$ bound state) production at $e^+ e^-$ collider (see [1] for instance), the $q\bar{q}$ pair produced by the $e^+ e^-$ scattering can be either in the singlet state or the octet state. The singlet is $\psi_s = \frac{1}{\sqrt{3}}(r\bar{r} + g\bar{g} + b\bar{b})$ while the 8 octets states are $\psi_{1…8}= r\bar{g}, g\bar{r}, r\bar{b}, b\bar{r}, g\bar{b}, b\bar{g}, \frac{1}{\sqrt{2}}(r\bar{r}-g\bar{g}), \frac{1}{\sqrt{6}}(r\bar{r}+g\bar{g}-2b\bar{b})$.
The $q\bar{q}$ pair must not carry hypercharge color number or isospin color number (but can however be member of octet). In other words it must be $r\bar{r}$ or $g\bar{g}$ or $b\bar{b}$. Hence, the total cross-section results from:  
$$ \sigma = \Sigma_{c\bar{c} = r\bar{r}, g\bar{g}, b\bar{b}} ~~|\mathcal{M}|^2 ~~\left(|<\psi_s|c\bar{c}>|^2 + \Sigma_{a=1..8}|<\psi_a|c\bar{c}>|^2\right)$$
where $\mathcal{M}$ is the amplitude for the electroweak process (same as the one giving the di-muons neglecting masses and taking into account the electric charge of quarks). It gives:
$$ \sigma = |\mathcal{M}|^2 ~~\left(\Sigma_{c\bar{c} = r\bar{r}, g\bar{g}, b\bar{b}}|<\psi_s|c\bar{c}>|^2 + \Sigma_{c\bar{c} = r\bar{r}, g\bar{g}, b\bar{b}}\Sigma_{a=1..8}|<\psi_a|c\bar{c}>|^2\right) = |\mathcal{M}|^2 (1+2)$$
Hence the famous factor 3. As mentioned by JeffDror, if the $q\bar{q}$ is produced in the octet state, since the initial state $e^+ e^-$ is a singlet, during the hadronization of $q\bar{q}$, there will be emission of gluons/quark pair to produce a singlet final state.
[1] http://cds.cern.ch/record/256588/files/P00018653.pdf
