Why is the electrical field outside spherical (shell) conductor radial if the charge $q$ inside is not concentric with the shell?

Then inside the shell the electrical field should be 0. Why is that?

The reason comes from your own assumption, which is that

Suppose we have a spherical shell that is a conductor

A conductor, if you have an electric field inside it, then charges will flow. After some time, the flow will cancel out, in any electrostatics situation. There can be an arbitrarily complicated pattern of surface charges, but they will act to cancel out any initially applied electric field inside a conductor.

Were you not already reading Griffiths? It covered this quite ok. Should not be a problem for you.

I'm thinking that the electrical field of the charge gets applied inside the shell in such a way positive charges are pushed to the outer surface and negative charges are drawn to the inside of the shell and thus a electrical field created by the redistribution of both negative and positive charges inside the shell neutralizes the electrical field of the charge q inside the shell.

This is correct.

Applying this to the shell the outer positive charges should not be uniformly distributed. There should be more negative charge on the inner surface closer to the charge and opposite to the negative charges here there should be more positive charges.

This is you misunderstanding what is happening.

The shell has a thickness, and an inner surface, and an outer surface. The inner surface will have its charge distribution quite complicated because the charge $+q$ is not placed at the one and only symmetry point.

But the inner surface will require $-q$ worth of charge to cancel out the charge as seen in the bulk of the shell, i.e. to cancel out the electric field inside the shell.

The outer surface is just receiving the $+q$ worth of excess charges and being told to distribute it. There is then all the freedom for the outer surface to distribute the $+q$ worth of charge in a spherically symmetric fashion, and that is what it will do.

i.e. inner surface and outer surface are essentially independently charged

This seems to be the crux of your confusion.

According to some explanations I've seen the negative charges alone on the inner surface neutralizes the electrical field of the charge inside such that the electrical field in conductor becomes 0. And the only way for the positive charges not to effect the field inside is by being distributed uniformly on the outer shell.

This is completely correct.

How can the negative charges alone make the field inside the conductor 0? If we take a small incision of the shell should it not be the same as the case with the rectangle.

No, you are wrong. There is a way to do this explicitly. However, it is of graduate level difficulty. But also luckily, it is considered simple enough that it is the first 3 chapters of the famous JD Jackson Classical Electrodynamics textbook.

Suppose we have a charge q at random inside a conducting cube. Will positive charges still be homogenously distributed on the outer side or is this just the case with a sphere because of symmetry?

The cube will fail to have spherical symmetry and thus will not have homogenous distribution. The cube tips will have more charges on them.

The simplest way to understand this is to recognize that the outer surface of the shell is an equipotential surface, for the reason you mentioned: that the shell is a conductor. If it wasn't equipotential, the charges on it would feel a force, and therefore they would move until equilibrium was reached.

If we know that the surface is at a constant potential and that the potential is $0$ infinitely far away, the solution to the laplace equation will be radially symmetric about the center of the sphere.