What is opposite to $\mathbf{w}_\parallel$ in a FBD of a box on a ramp? I tried doing research on this but to no avail so my question is this:

If the normal force of an object with mass $m$ on a ramp inclined with angle $0<\theta<90^\circ$ is equal and opposite to the component of gravity pulling the object perpendicularlly into the ramp ($\mathbf{w}_\perp$), then what force is "equal and opposite" to the component of gravity that is parallel to the ramp?

As newtons third law says, there has to be an equal and opposite force of $\mathbf{F}_w$.
$\mathbf{F}_\perp$ seems to only take care of the perpendicular component $\mathbf{w}_\perp$, so what takes care of the parallel component $\mathbf{w}_\parallel$? Is it friction?
 A: 
As newtons third law says, there has to be an equal and opposite force of Fw.

Indeed, but your interpreting it wrong. Put it this way,
The force on A due to B is equal and opposite to the force on B due to A. The key point being that the action-reaction pairs are forces on different objects. Not the same object. (the box)
Let us identify the action reaction pairs -

*

*The normal force on the box by the ramp is at an inclined angle, perpendicular to the ramp. And the normal force on the ramp by the box exactly opposite in direction, inclined downwards.

*The gravitational force on the box by the earth which points straight down. And the gravitational force on the earth by the box which points straight up.

So the normal force on the box is not an action-reaction pair with gravity.
To put it inline with your question,

*

*Normal force perpendicular to the ramp felt by the box is equal and opposite to the normal force experienced by the ramp due to the box.

*The gravitational force perpendicular to the ramp is equal and opposite to that component of gravitational force felt by the earth.

