Torque and force Does the force due to torque on the edge of a wheel depend on the mass of the the wheel or is it always $\tau / R$?
 A: The torque due to a force is not dependent on the mass of the object it is acting on. It is dependent on the point you are taking the torque about. In general, the torque due to a force is the cross product of the position vector from the point you are taking the torque about to the point of application of the force and the force. 
If your drawing is intended to represent a system of a force and a torque acting on the wheel, then in order to be in equilibrium the force should be pointing in the opposite direction. The drawing is correct if it is simply indicating the torque about the center due to the force acting tangentially to the rim of the wheel. 
EDIT: Since you appear to be asking about a larger system, e.g. wheels attached to a car with torque supplied to the wheels by an engine, then this diagram and highly simplified analysis will perhaps better address your question. $m$ is the mass of the vehicle, $L_1$ and $L_2$ are distances from the front and rear wheels to the center of mass of the vehicle, $R$ is the radius of the wheels, $\tau$ is the torque supplied by the engine, $N_1$ and $N_2$ are the normal forces that the ground imparts on the wheels, and $F$ is the friction force on the wheel being driven by the engine (front wheel drive is being assumed here).

The car has no angular acceleration and let's assume we can neglect the rotational inertia of the wheels for the overall analysis of the angular momentum of the vehicle. Also note that the torque supplied by the engine to the vehicle is internal to the system and hence there is a balancing torque in the opposite direction from the car on the engine. 
So then the sum of the external torques on the vehicle about the center of mass of the vehicle is zero.
$-N_1 L_1+ N_2 L_2 -F h = 0$ 
where $h$ is the vertical distance from the ground to the center of mass of the vehicle. 
There is no acceleration of the system in the vertical direction, so the sum of the forces in the vertical direction is zero.
$N_1+N_2-mg=0$
The forces in the horizontal direction lead to the acceleration of the vehicle in the horizontal direction $a$.
$F = ma$
Finally, if we analyze the front wheel on its own, then we do need to consider its rotational inertia.  We also note that the forces from the axel pass through the center of the wheel.  Then, the angular acceleration of the wheel is given by:
$\tau - FR = I \alpha$
where $I$ is the moment of inertia of the wheel about its center and $\alpha$ is the angular acceleration of the wheel.  If the wheel does not skid then the acceleration of the vehicle is related to the angular acceleration of the wheel as:
$a=\alpha R$
If you know the torque supplied $\tau$ then you can use these five equations to determine the normal forces on the wheels, $N_1$ and $N_2$, the friction force with the ground $F$, the acceleration of the vehicle $a$, and the angular acceleration of the wheels $\alpha$. 
The force you were asking for is then:
$F=\tau/(R + I/mR)$
So, if the mass of the car times the radius of the wheels is much greater than the moment of inertia of the wheels, which is generally the case, then the force $F$ is well approximated by $\tau/R$.
If the wheels skid, then you need to introduce the coefficient of kinetic friction to the problem.
A: Force due to torque will not depend on the mass of the wheel.
If the mass of the wheel is  more, more amount of work needs to be done on the wheel. But we are not caring about that. 
Remember here we have produced the torque ( tau) ( that is the initial condition provided to us), now your formula will always be valid.
A: A car engine puts effort into producing a torque about the wheel's axle. This torque makes the wheel turn.
If the wheel is heavy (large mass, thus large moment of inertia), then it is harder for the engine to produce this torque. 
But if the engine succeeds in producing the torque, then that torque causes a force at the road (at the contact point between road and wheel). And the relation is:
$$\tau=rF_\perp$$
This expression always applies. And it does not involve the wheel mass. That mass influenced the effort required to produce the torque, but not the force caused by that torque.
The answer to your question is no.

Note that in other situations you might think og the torque being produced by the force. Such as when turning a bolt with a wrench. Then the situation is reversed. Then it takes effort to produce the force (effort by your hand for example). And the heavier the wheel, the larger the necessary effort.
Still, in this situation as well, the torque that this force causes does not depend on the mass, but again follows the expression above. 
A: Yes, it does depend on wheel mass, but indirectly. Here is how =>
By definition torque is:
$$ \tau = r\,F_\perp $$
And Newton second law expressed for rotation is :
$$ \tau=I\,\alpha $$
where $I$ is moment of inertia and $\alpha$ is angular acceleration. Putting these equations together and solving for a force gives
$$ F_\perp = \frac{I\,\alpha}{r} $$
In your case, wheel can be seen as a solid disk, which has moment of inertia:
$$ I_\textrm{disk} = \frac{m\,r^2}{2} $$
Substituting moment of inertia back into force equation gives final solution :
$$ F_\perp = \frac{m\,r\,\alpha}{2} $$
So you need wheel mass $m$, radius $r$ and angular acceleration $\alpha$ to calculate resulting force. However, if you have already calculated/extracted wheel's torque, then you can calculate same thing by just $\tau\,/\,r$. Still you need to remember that torque according to second Newton law depends on rotating object moment of inertia. And moment of inertia depends on an object mass AND on object shape. You can look for common shapes moment of inertia here. 
In a general case if body is continuous, then moment of inertia can be found integrating infinitesimal masses in a body:
$$ I = \int r^2 \,dm $$
EDIT :
If you need to include ground resistance effect on wheel, then just decouple net force :
$$ F_\perp = F_{\perp\text{engine}} - F_{\perp\text{resistance}} 
\\= F_{\perp\text{engine}} - \mu_{s}\,m\,g
$$
