Why do charges on the sphere not fly? Consider a uniformly charged conducting solid sphere. As we know, all of the charge must reside on the surface of the sphere, so let's only take this uniformly charged spherical plane into consideration.
I have color-labelled six representative charge elements ($dq$) on this spherical plane.
Let's take a loot at the red charge element.
The electric field just outside the red charge element (outside the plane) is $\sigma/\epsilon_0$ and that just inside the red charge element (inside the plane) is zero. But, what is the electric field at the red charge element?

Imagine we remove all other charge elements from this spherical plane except the purple charge element (without allowing the purple charge element to change its position on the plane) and compute the field contributed by this purple element at the red dot. If we repeat this process for all of the five non-red charge elements and summate their fields at the red point, we find that the field at the red charge element must be non-zero. So, why doesn't the red charge element just fly off?
 A: Standard answer for a discountinous field is to take the average of the inside and outside field. So in your case the field on the sphere would be $\sigma/(2\epsilon_0)$. See for example Purcells Electricity and Magnetism. 
For your second question: In a conductor the electic charge is only allowed to flow inside the conductor. Normally charge does not flow off the conductor (except if you heat it up a bit); in that case you would loose charge as you mentioned.
A: The electric field on the surface of the conductor is the average of the field above and below the surface. This can be seen as follows: Consider a patch of charge on the surface. The total field is the sum of the field ($E_{patch}$) due to the patch and field due to the rest of the charges not included in the patch ($E_{other}$). So, field just outside the surface is: $E_{above}$ = $E_{other}$ + $\sigma/2\epsilon_o$[ this is the contribution of the patch].
Similarly $E_{below}$= $E_{other}$ - $\sigma/2\epsilon_o$. So, it can be seen easily that the $E_{other}$ is the average of field above and below. Now the force on the patch is again due to these two field[patch and other]. Since the patch cannot exert a force on itself, the force on the patch is only due to $E_{other}$. The field is defined as force per charge, so the field on the surface on the surface is the average (=$E_{other}$). This is true for any patch and hence for the entire surface. 
To answer your second question, the charge cannot jump off due its own field [the charge cannot experience it's own field]. It does experience a force due to other charges, hence the surface is under a pressure.     
