Not that high a force is necessary. It depends on the angle of the incline. The steeper the incline, the greater must be the force. Identify the forces acting on the box, and see how you can incorporate the angle of the incline into your calculation.
I have figure that out, and the only force acting, in this case, is gravity. So that would mean I would need a force mgx in the opposite direction but slightly greater correct? (mgx is the x-component of mg)
Yes, where the x component of gravity is the component acting down and parallel to the incline ( the incline is chosen as the x axis). Note in your first post you mentioned the box has a weight of m. It has a mass of m and a weight of mg.
Yes, slightly greater the component of the weight, that is, slightly greater than mg(sin theta), where theta is the angle that the incline makes above the horizontal. We say here "slightly greater" because presumably it starts from rest, so you have to apply a slightly greater force to get it moving by accelerating it to some small speed, and once it is in motion you merely have to apply exactly mg sin theta force to keep it moving at that constant speed. This of course assumes no friction or retarding forces are acting, just gravity force down the plane.
No, it is already moving! An object at rest or in motion at constant speed in a straight line will remain at rest or in motion at constant speed in a straight line if the net force is 0. Per Mr. Newton.
Well you could let go of it and it would come to a temporary stop before reversing direction and accelerating down the incline to a greater and greater speed. But if you want to bring it to a permanent rest, you have to apply a force slightly less than mg sin theta to stop it ( decelerate it to zero speed), then once it stops apply exactly mg sin theta to keep it at rest.