One way to see a rigid body is a collection of infinity particles such that they preserve the distances between themselves (can only translate and rotate). So, we have a constraint, and then internal forces that maintain this constraint.
This internal forces are dependent on the forces that you apply in the system. If you apply a force in one point of the object (one particle), then all other points would feel a force such that the accelerations of the hole system don't destroy the constraint. Rigid bodies need to have energy stored there to do that.
If you have a static homogeneous rigid body with the center of mass $\vec{r}_{cm}=\vec{0}$, applying a force $\vec{f}(\vec{r}_0)$ in some point $\vec{r}_0$ such that the force is paralel to $\vec{r}_0$, then all the points of the body need to feel this same force to preserve the constraint. The result of this force is an aceleration of the whole system generating a translation.
If the force $\vec{f}(\vec{r}_0)$ is not parallel to $\vec{r}_0$, then all the points need to feel a force $\vec{f}(\vec{r})$ such that
$$
\vec{r}\times\vec{f}(\vec{r})=\vec{r}_0\times\vec{f}(\vec{r}_0)
$$.
because the constraint is mathematically translated to $d\vec{r}=d\vec{\theta}_0\times\vec{r}$ for every $\vec{r}$ and the same $d\vec{\theta}_0$ (see here for understand why).
With this we can define a quantity called Torque associated for a force $f$ applied to a point $\vec{r}$ of the rigid body by:
$$
\vec{\tau}=\vec{r}\times\vec{f}(\vec{r})
$$
So, if we have a force $\vec{f}_1$ applied at $\vec{r}_1$ and $\vec{f}_2$ applied at $\vec{r}_2$ we can see that the net force at point $\vec{r}$ need to obey:
$$
\vec{f}(\vec{r})\times\vec{r}=\vec{f}_1\times\vec{r}_1+\vec{f}_2\times\vec{r}_2
$$
Then we may define a net torque:
$$
\vec{\tau}_{net}=\sum_{j}\vec{f}_{j}\times\vec{r}_{j}
$$
And, if $\vec{\tau}_{net}=\vec{0}$, then the body does not rotate because at each point $\vec{f}(\vec{r})\times\vec{r}=0$ and for $\vec{r}\neq \vec{0}=\vec{r}_{cm}$ we conclude that the force $\vec{f}(\vec{r})$ is paralell to $\vec{r}$.
Newton's second law originally assumed that the mass was a constant of nature, at least if you write it as F=dp/dt. It will only work with a changing mass if that mass leaves the body at the same speed than the original object. To understand why, just think you have a composite object moving a constant speed. If you now only watch at one half of the object, the mass will be reduced to half, but the speed will stay constant (we are assuming no internal forces so both halves keep moving at the same speed. Now, if both halves interact so that the one at the "front" pushed the one at the back apart ("a digital one step fluid"), you will have an interaction between the two halves, and the right way to describe it is to use the initial speeds and the interaction. Or using that the total moment is a constant, but always considering the mass of each subpart as constant. If you just use the second law with the derivative of the mass, you will get a different (and incorrect) result.
Your last equation for rigid bodies is incorrect (in the sense of non-physical). The correct one is:
$\frac{\text{d}\mathbf{Q}}{\text{d}t} = m\mathbf{\dot{v}} =\sum_i \mathbf{F_i^\text{ext}}$
Because in Newtonian mechanics mass is a constant for a rigid body. I do not have any references beyond my professor telling me that and solving problems (such as the rocket) in both ways and getting different results, with the result using non-variable mass being the correct one. Just think that in nature there is no classical mechanism that allows a rigid body to loose or change mass (unless it is composite and losses some parts). Now, the change of mass due to relativity theory is correct, but Newton laws no longer apply in that case.
Just a funny note: Both Plastino's father and Muzzio were professors of mine!
Best Answer
The internal forces come in equal-and-opposite pairs (Newton's 3rd), and therefore result in no net force on the object. If you did take the vector sum of all of them, they would just cancel out. Including them therefore doesn't change the final expression.
There is no need for idealization - just the acceptance of classical mechanics.
Note that this is even true for non-rigid bodies: while the position of their center of mass will change, that center of mass will still move in accordance with the vector sum of the external forces.