Texts on QCD don't divide the generators of $SU(3)$ – and therefore "bicolors of gluons" – into two groups because this separation is completely unphysical and mathematically artificial (basis-dependent).
Moreover, the number of "bicolors of gluons" i.e. generators of $SU(3)$, the gauge group of QCD, isn't nine as you seem to think but only eight. The group $U(3)$ has nine generators but $SU(3)$ is the subgroup of matrices with the unit determinant so one generator is removed. At the level of gluons or Lie algebra generators, the special condition $S$ means that the trace is zero. So the combinations
$$ A(r\bar r) + B (g\bar g) + C(b \bar b)$$
are only allowed "bicolors of gluons" if $A+B+C=0$. Now, in this 8-dimensional space of "bicolors of gluons", there are no directions ("bicolors") that are better than others. For any direction in this space, there exists an $SU(3)$ transformation that transforms this direction into a direction non-orthogonal to any chosen direction you choose. This is true because the 8-dimensional representation is an irreducible one (the adjective "irreducible" means that one shouldn't try to split it to two or several separated collections!). And there doesn't exist any consistent Yang-Mills theory that would only contain the six off-diagonal "bicolors" because the corresponding six generators aren't closed under the commutator.
The actual calculations of the processes with virtual gluons ("forces" between quarks etc.) therefore never divide terms to your two types because this separation is just an artifact of your not having learned group theory. Instead, all the expressions are summing over three colors of quarks, $i=1,2,3$ indices of some kind, and there is never any condition $i\neq j$ in the sums because such a condition would break the $SU(3)$ symmetry.
Now, the $r\bar r,g\bar g,b\bar b$ "bicolors of gluons" (only two combinations of the three are allowed) are actually closer to the photons than the mixed colors. So it's these bicolors that produce an attractive force of a very similar kind as photons – they are generators of the $U(1)^2$ "Cartan subalgebra" of the $SU(3)$ group and each $U(1)$ behaves like electromagnetism. That's why these components of gluons cause attraction between opposite-sign charges and repulsion between the like charges.
The six off-diagonal "bicolors of gluons" (and let me repeat that the actual formulae for the interactions never separate them from the rest – they're included in the same color-agreement-blind sum over color indices) cause neither attraction nor repulsion: they change the colors of the interacting quarks so the color labels of the initial and final states are different. It makes no sense to compare them, with the idea that only momentum changes, because that would be comparing apples and oranges (whether the force looks attractive or not depends on the relative phases of the amplitudes for the different color arrangements of the quarks).
At any rate, particles like protons contain quarks of colors that are "different from each other" so they're closer to the opposite-sign charges and one mostly gets attraction. However, the situation is more complicated than it is for the photons and electromagnetism because of the six off-diagonal components of the gluons; and because gluons are charged themselves so the theory including just them is nonlinear i.e. interacting.
Most generally, the force $\vec F_m$ and torque $\vec \tau$ due to the magnetic field $\vec B$ acting on a particle with a charge $q$ and magnetic moment $\vec\mu$, moving with velocity $\vec v$ can be written as
\begin{align}
\vec F_m &= q \vec v \times \vec B + \vec\nabla (\vec B\cdot\vec\mu),
&&&&(1)\\
\vec\tau &= \vec r \times \vec F_m + \vec\mu \times \vec B. &&&&(2)
\end{align}
The first term in (1) is the standard Lorenz term, while the latter is usually not mentioned as a fundamental law since classical electrodynamics did not assume point-like particles with spin and magnetic moment but rather modeled magnetic dipoles by an electric current loop, behaviour of which in $\vec B$ can be derived just from the first term (focres act on parts of a the electric circuit because there are moving charges inside).
In Eq. (2), the first term $\vec r \times \vec F$ is the usual textbook definition of a torque in the case of a force acting on a point particle - it is just a tool to understand how will the tracejtory of the particle be curved due to $\vec F_m$ (i.e., it tells us something about orbital motion).
Apparently, it depends on the choice of your coordinates.
In contrary (but analogously to the previous paragraph), the second term in Eq. (2), $\vec \mu \times \vec B$, is an entirely new piece of physical law for a point-like particle. It is not present usual chapters on classical mechanics since they do not assume that that particles have spin (it is hard to imagine a rotating point-like particle). Again, if you modeled magnetic dipoles by an electric-current loop, you can derive the second term in (2) from $\vec F_m = q \vec v \times \vec B$.
However, this still does not answer the core of my question here which is, even so with this applied torque mentioned, why is then a free electron as suggested by known physics, under this torque not forced to any translational motion (i.e. orbital or linear)?
Now, let us have a look at what forces and torques actually do: Changes in motion of the center-of-mass of a body are always determined solely by the force, according to the famous Newton's law $\vec F = m \vec a$. Denoting the body's momentum by $\vec p=m\vec v$, this law can be also reformulated as $$\vec F = \frac{\mathrm d\vec p }{\mathrm dt}. \qquad (3)$$
In contrary, torque causes changes in angular momentum $\vec J$ (i.e., in "rotation"):
$$ \vec \tau = \frac{\mathrm d \vec J}{\mathrm d t}. \qquad (4)$$
Generally, angular momentum of a moving particle with spin is a sum of two terms, $$\vec J = \vec L + \vec S$$ where $\vec L = \vec r \times \vec p = m\vec r \times \vec v$ is the usual orbital angular momentum while $\vec S$ is the particle's internal angular momentum, i.e., spin.
Now, if $\vec F_m = 0$, Eq. (3) tells us that there will be no acceleration. This is perfectly consistent with Eq. (4) which yield
$$
\vec \tau = \frac{\mathrm d\vec J}{\mathrm dt}
= \frac{\mathrm d\vec L}{\mathrm dt} + \frac{\mathrm d\vec S}{\mathrm dt}
= \frac{\mathrm d(\vec r \times \vec p)}{\mathrm dt} + \frac{\mathrm d\vec S}{\mathrm dt}
=\vec v \times (m\vec v) + \vec r \times \vec F + \frac{\mathrm d\vec S}{\mathrm dt}
=0+0+\frac{\mathrm d\vec S}{\mathrm dt}.
$$
I.e., in the absence of a magnetic force, the magnetic torque acts solely on the spin of the particle. Using Eq. (2):
$$\vec \mu \times \vec B =\vec \tau = \frac{\mathrm d\vec S}{\mathrm dt}. $$
Btw., notice that since the particle's magnetic moment $\vec \mu$ is proportional to its spin $\vec S$, the torque is always perpendicular to $\vec S$ and hence may cause a change in the orientation of the spin (i.e. precession) but not of its size (i.e. electrons always have spin of 1/2 $\hbar$ - but let's not dig into the quantum world in this discussion).
Is this above paragraph actually true or just never experimentally or analytically verified up to now?
It is true as far as classical (non-quantum) electrodynamics is true. As noted by others in the comments, a theory cannot be verified but only disproved. Electrodynamics is the most precisely measured part of physics whatsoever.
- What happens when I approach the pole of a permanent magnet (non-homogeneous field) close to a copper rod? Will the free electrons inside the copper rod initiate an orbital motion besides there known gyromagnetic motion under the influence of this torque applied by the external magnetic field, opposite to the orbital direction of the unpaired electrons inside the permanent magnet and therefore generate also the diamagnetic property of the copper material?
Yes, diamagnetism is a property of all materials. But notice that the origin of the torque causing the orbital motion is not the magnetic moment of the electron but the first term, $\vec r \times \vec F_m.$
- What about a free stationary electron inside a vacuum environment? Will it start to move towards a magnet even if it was initially stationary under the influence of a non-homogeneous magnetic field?
Yes.
Will the same behavior occur with a homogeneous magnetic field and if not why?
No. For a non-moving electron in a homogeneous magnetic field, both terms in Eq. (1) are zero. The electron might only undergo spin precession.
If this is the case for the free electrons in the copper rod or the vacuum is this then not an actual prove that electrons DO move translational (free electrons forced to orbital motion besides their gyromagnetic motion) under a static magnetic force even if they were initially stationary?
As explained above, this does not happen in vaccum. In the material, we must be careful about what we mean by stationary. In the quantum language, an electron in its atomic orbital is stationary; however, its classical analogue is that it is somehow circling around the nucleus and hence it undergoes an orbital motion eve initially before you apply the external magnetic field. Hence, the answer is NO.
Is it possible and is this ever experimentally verified, given a very strong external magnetic field so that the arc trajectory of this orbital motion under the applied torque the magnetic moments of the free electrons are subjected with, to be so large that it will force a free electron practically in a linear motion?
Are you asking wheather we can compensate the curving effect of the magnetic force $q\vec v \times \vec B$ by the interaction of the electron's magnetic dipole $\mu$? I think the answer is NO, since both effects are directly proportional to the same field $\vec B$.
Finally, does not the above mentioned, leads in the inescapable conclusion that magnetism is not an exclusive emergent (side effect) phenomenon of electron charge translational motion but actually originates from the intrinsic property of spin magnetic dipole moment of the electron similar as electric phenomenon originates from the intrinsic property of the charge of the electron?
Indeed. If you accept the semi-classical picture of a point-like electron with non-zero spin and magnetic moment, then there are two sources of magnetism (moving charges and particle's spin), which also both respond to an external magnetic field (first and second terms in Eqs. (1) and (2).)
Nevertheless, note that in relativistic quantum theories, both "sources of magnetism" naturally emerge from a single physical law (i.e. Lagrangian of quantum electrodynamics).
Therefore the origin of the electromagnetism phenomenon in general must exclusively be attributed to the whatever unknown intrinsic mechanics of the electron itself?
No, not all magnetic phenomena are caused by the magnetic dipoles of elementary particles - see the first terms in Eqs. (1) and (2). Note that pions are charged particles without spin or magnetic moment but their trajectories can still be bent by magnetic fields (and it is an experimental routine).
Best Answer
In order to understand asymptotic freedom, you need to be aware of the concept of renormalization. Since you want a qualitative description, just think of renormalization a modification of the coupling strengths and masses of particles at high energies. This is roughly like pushing a ball through the water; the harder you push, the more the water sticks around it and the harder it is to move. This can be modeled with Newton's 2nd law $F=ma$ by replacing the mass with a slightly larger mass $m+\delta m$, and this $\delta m$ depends on the velocity of the ball in the water.
(that discussion can be found in section 3.2 of Connes and Marcolli, "Noncommutative Geometry, Quantum Fields and Motives")
Once you have the concept of renormalization, asymptotic freedom is a property the strong force has as you scale the coupling constant to high energy. Rather then the coupling getting stronger, it gets weaker. This has major consequences for confinement - that is, bound quarks. At low energies, quarks in bound states are forever bound - it becomes harder and harder to pull them apart the further apart you pull them. At high enough energies (say, colliding two protons at 7 TeV like the LHC) the quark coupling gets small and quarks are essentially free and unbound. It should be easy to see how this would change the cross section.
As a sidenote, only the strong force is asymptotically free. The E/M and weak force become stronger as the energy gets higher. In addition, it is important to realize that we cannot solve problems involving the strong force at low energies (if you could, the Clay Mathematics Institute would give you $1 million!). Once they are at high energies, the strong coupling is weak so QCD acts quite a bit like QED.