Does $F=ma$ still apply when a force creates torque? Consider a wheel and axle. I apply a force on the edge. That creates a torque. However, why doesn't $F=ma$ still apply? Aren't I still applying a force to the entire wheel/axle assembly, why doesn't it accelerate linearly?
Is it because the axle supplies a reaction force such that there is no net force on the wheel axle assembly, just a torque.
My question exists because of how a cars tire works, where friction from road both creates a counter torque and a push forward. However, many times a force just creates a torque. 
 A: Almost exactly right.  
The axle provides a force in the opposite direction that's equal and opposite to your applied force.  But it's not a reaction force to your force.  It's a separate force. An analogy would be a book lying on a table.  The book applies a force to the table.  What keeps the table from accelerating?  The normal force of the floor on the legs of the table.  That force is not a reaction force to the force the book applies to the table.  The reaction force of the book on the table is the normal force of the table on the book.
If the wheel were unconstrained (out is space somewhere) then $F=ma$ would apply as you imagine.  There would be linear acceleration and angular acceleration.
A: The relationship $\boldsymbol{F} = m\, \boldsymbol{a}_{\rm CM}$ always holds true as long the acceleration is that of the center of mass. The is true because the definition of momentum is $\boldsymbol{p} = m\, \boldsymbol{v}_{\rm CM}$, and force is the rate of change of momentum.
When a force is applied away from the center of mass (say point A), not only the center of mass accelerated by $\boldsymbol{a}_{\rm CM}$, but also the body has rotational acceleration $\boldsymbol{\alpha}$.
The rotational acceleration is a result of the torque $\boldsymbol{r}_A \times \boldsymbol{F} = \mathcal{I}_{\rm CM} \boldsymbol{\alpha}$ where $\boldsymbol{r}_A$ is the location of point A relative to the center of mass. Here $\mathcal{I}_{\rm CM}$ is the mass moment of inertia tensor at the center of mass.
If the body is initially at rest, the linear acceleration of point A is $\boldsymbol{a}_A = \boldsymbol{a}_{\rm CM} + \boldsymbol{\alpha} \times \boldsymbol{r} $ 
So now you can find the effective mass of point A by defining 
$ \boldsymbol{F} = m_{\rm eff}\, \boldsymbol{a}_A $, and using the above equations
$$ \boldsymbol{F} = m_{\rm eff}\, ( \boldsymbol{a}_{\rm CM} + \boldsymbol{\alpha} \times \boldsymbol{r}_A) = m_{\rm eff} \left( \tfrac{\boldsymbol{F}}{m} + \mathcal{I}_{\rm CM}^{-1} ( \boldsymbol{r}_A \times \boldsymbol{F}) \times \boldsymbol{r}_A  \right)$$
In the 2D example of a tire, with $\boldsymbol{F} = \pmatrix{F \\ 0 \\0}$ and $\boldsymbol{r}_A = \pmatrix{0 \\ -R \\ 0}$ the above equation becomes
$$ F = m_{\rm eff} \left( \tfrac{F}{m} + \tfrac{F R^2}{ I_{\rm CM}} \right) $$ or $$ \boxed{ m_{\rm eff} = \frac{1}{ \tfrac{1}{m} + \tfrac{R^2}{I_{\rm CM}} } } $$
where $I_{\rm CM}$ is the component of $\mathcal{I}_{\rm CM}$ along the z-axis (rotation axis).
If the wheel was a uniform disk, then $I_{\rm CM} = \tfrac{m}{2} R^2$, and the effective mass is $m_{\rm eff} = \tfrac{1}{3} m $. This means the force feels only 1/3 of the total mass as the point A under the force accelerates more than the center of mass does.
A: In the case of extended bodies $\mathbf F = m \mathbf a$ applies only to the centre of mass. Other points experience net torque which causes them to gain additional angular acceleration hence for such particles $\mathbf F \neq m \mathbf a$ 
A: If the force is directly in line with the center of mass of any unrestrained object, it will cause linear acceleration only, if the force is not directly in line with the center of mass, it will cause linear acceleration and rotation. Consider the book on the table, if you push it away from you on the center of an edge of the book (in line with it's com), it will slide across the table. If you push on it more towards a corner (not in line with it's com) it will rotate and move away.
