I don't really like summaries like that cause there's too many short-cuts taken in the summary - but that's my personal take, you don't have to share that opinion. I like this one better, as it has pictures:
http://abyss.uoregon.edu/~js/cosmo/lectures/lec22.html
The first sentence you quoted - as pointed out, is incorrect, or, at least, said badly.
"As the universe expanded further, and thus cooled, common particles began to form. These particles are called baryons and include photons, neutrinos, electrons and quarks would become the building blocks of matter and life as we know it. "
I'm willing to guess, what he/she meant to say these "common particles" included Baryons, Photons, Neutrinos, Electrons and Quarks", cause to say Baryons include photons is incorrect.
It might help to look at the standard model as there's 3 main categories: Quarks, Bosons and Leprechauns, I mean, Leptons (just checking to see if you're paying attention). Quarks form Baryons and Mesons, (there's probobly a Free Meson joke to be made, but I'll try to resist). Leptons include the very useful Electron as well as Neutrinos and Bosons are essentially binding particles, and/or energy transfer particles - also very useful. The photon is a Boson. The Higgs field creates a Boson too, but that's a little more complicated and probobly best left for later. Anyway - a glance at the standard model will help with the 3 categories:
Standard Model: http://upload.wikimedia.org/wikipedia/commons/thumb/0/00/Standard_Model_of_Elementary_Particles.svg/2000px-Standard_Model_of_Elementary_Particles.svg.png
Particles are defined by Charge, Spin and Mass, and quarks also by color, and there's also anti particles, which, when an antiparticle and a particle meet, they annihilate in a flash of heat/light - so the early universe was very very hot and heat tends to prevent bonds from forming, even very small, very strong bonds inside an atom - but as the universe cooled, bonds began to form and that meant, Protons and Neutrons and later, some simple nuclei like deuterium and helium and later, when it cooled enough Electrons could bind with Nuclei and you had atoms.
When your article says:
"Although lighter particles, called leptons, also existed, they were prohibited from reacting with the hadrons to form more complex states of matter. These leptons, which include electrons, neutrinos and photons, would soon be able to join their hadron kin in a union that would define present-day common matter."
Well, first, Photons aren't Leptons at all. Second, Neutrinos very rarely react with Hadrons (and it's probobly better to use the word Baryon in this sense than Hadron, Baryons are a class of Hadron). - see link below:
http://www.particleadventure.org/hadrons.html
I think a much simpler way to say to look at all of this, as I said above, is that heat breaks bonds. This is pretty much universally true in all of physics and chemistry, where as, the forming of bonds releases heat.
Very hot universe - you get "quark soup" - which isn't just quarks, there's also Leptons and Bosons, but Quark soup is a term some use and I rather like it.
Less hot universe, you get Protons and Neutrons (Baryons) also, Leptons (electrons) and Bosons, and it's not true that Baryons and Electrons don't interact at this point. They interact plenty, but what they don't do is form stable atoms.
a bit less hot than that, some of the Protons and Neutrons bind and you get 2 or more Baryons bound into an Atomic Nuclei, (technically a Proton is an atomic Nuclei all by itself, but as the universe cools enough you get some Deuterium nuclei and Helium nuclei and a very small amount of Lithium and Beryllium Nuclei are all formed - then, as it cools a bit more, this creation of Nuclei stops. It's possible to calculate how much hydrogen Nuclei vs Helium Nuclei there should be and the fact that the universe matches this calculation very closely is a good evidence that the big-bang and parts of the standard model theories are mostly correct. This all happens pretty quickly, both the period where helium and a few other nuclei begin to form and when they stop forming.
It's not yet what we think of as classical matter, just the a few Nuclei. Stray Neutrons aren't stable, so they either bind with Protons to form atomic Nuclei or decay into a Proton, an Electron and other smaller stuff. The entire universe is still crazy hot. Cooled down significantly, but still hot.
After a significantly longer amount of time (380,000 years or so) it's cool enough for Electrons to bind to Atomic Nuclei and this is when you get what we recognize as standard matter - hydrogen and Helium atoms instead of plasma, and when this happens, the universe becomes clear. Plasma is opaque, you can't see through it. It probobly looks something like smoke, but hydrogen and helium atoms are essentially transparent and at this point, the universe becomes transparent and light can travel through it without constantly bumping into a baryon, nuclei or electron and this is where the cosmic background radiation comes from - at the time, it was like a great big flash of light, well, X-ray light probobly, not visible and you wouldn't have wanted to be there was it was still 3,000 degrees. But the universe now has lots and lots and lots of atoms, which was pretty cool, as it was the beginning of what we'd recognize as a universe, not plasma or particle-soup.
It continued to expand and cool and gravity and dark matter helped pull the zillions of loose atoms together into galaxies and stars and all that good stuff.
It's entirely possible that a number of stars formed before cosmic background radiation event, cause stars can form from hot plasma, so, I would think some stars had already formed but we can't see any evidence of that cause all light back then was reflected every which way cause the universe wasn't transparent. We can only see back to the cosmic background event, but we can re-create what particle interactions and study what might have happened in the hot young universe using a particle accelerator.
I welcome any correction to any of the above, but that's how I look at the early universe. I Love this stuff & think about it a lot.
Best Answer
Each quark $q$ and antiquark $\bar{q}$ transform in the fundamental representation ${\bf 3}$ and antifundamental representation $\bar{\bf 3}$ of the color group $SU(3)_C$, respectively.
However, a hadron has to be a color singlet ${\bf 1}$, due to color confinement.
In particular the center $$\mathbb{Z}_3~:=~\mathbb{Z}/3\mathbb{Z}~=~\{1,e^{\pm 2\pi i/3}\} ~\subseteq ~SU(3)_C$$ of the color group $SU(3)_C$ should be in a trivial representation of the cyclic group $\mathbb{Z}_3$. This precisely happens if the quark number is a multiple of 3. This answers OP's title question.
Examples:
A single quark $q$ transforms in the fundamental representation ${\bf 3}$ of $SU(3)_C$, and is hence not allowed. See also related Phys.SE post here.
A diquark $qq$ belongs to the tensor representation ${\bf 3}^{\otimes 2}:={\bf 3}\otimes{\bf 3}\cong\bar{\bf 3}\oplus{\bf 6}_S$, which we have decomposed in irreps of $SU(3)_C$. This contains no singlet ${\bf 1}$, and is hence not allowed. See also related Phys.SE posts here and here about $SU(3)$ tensor representations.
In a meson, the quark-antiquark-pair $q\bar{q}$ belongs to ${\bf 3}\otimes\bar{\bf 3}\cong{\bf 1}\oplus{\bf 8}_M$, which contains a singlet ${\bf 1}$, and is hence allowed.
The third tensor product is ${\bf 3}^{\otimes 3}={\bf 3}\otimes{\bf 3}\otimes{\bf 3}\cong {\bf 1}\oplus 2\cdot{\bf 8}_M\oplus{\bf 10}_S$. In a baryon, the three quarks $qqq$ form a totally antisymmetric representation $\wedge^3 {\bf 3}\cong {\bf 1}$ of $SU(3)_C$, which is isomorphic to a singlet ${\bf 1}$, and is hence allowed. See also related Phys.SE post here. The lightest baryon, the proton, is stable in the standard model due to baryon/quark number conservation. (However, see the hypothetical proton decay.)
The tensor product ${\bf 3}\otimes{\bf 3}\otimes\bar{\bf 3}\cong 2\cdot{\bf 3}\oplus {\bf 6}_M\oplus {\bf 15}_M$ contains no singlet ${\bf 1}$, so the combination $qq\bar{q}$ is not allowed.
The fourth tensor product is ${\bf 3}^{\otimes 4}\cong 3\cdot{\bf 3}\oplus 2\cdot{\bf 6}_M\oplus 3\cdot{\bf 15}_M\oplus{\bf 15}_S$. This contains no singlet ${\bf 1}$, so four quarks $qqqq$ are not allowed.
A "molecule" of mesons and baryons, such e.g. a tetraquark $q\bar{q}q\bar{q}$ or a pentaquark $qqqq\bar{q}$, is also allowed, but is obviously heavier. See also related Phys.SE posts here, here, and here.
TL;DR: The number of quarks minus the number of antiquarks should be divisible by 3.
References: