For generic $F$ the answer is probably yes, but for certain $F$ and certain choices of $\epsilon_i$ the answer is no, not necessarily. 

Here is an example with n=2 and $M$ the Klein bottle. We start with F being a standard projection of the Klein bottle into the plane. This is easy to visualize by googling for images of these glass Klein bottles that are everywhere. 

With the standard projection we get that $W_F$ is an embedded circle and $M_F$ has just one component. So in that case the integral will be non-zero for each of the two possible orientations. 

However now we are going to deform F slightly. We can imagine that our glass Klein bottle has a lump/boil/blister and that it is really quite large. This gives use a new projection from the Klein bottle to the plane. The image looks something like this:
[![a projection of a Klein bottle][1]][1] 

The new $W_F$ is the inverse image of the red curve. It is now two circles. $M_F$ now has two components, one of which is a disk. Topologically we have taken the old $M_F$ and removed a disk. 

When we integrate $\omega_F$ over these two pieces we will get numbers and by expanding the "boil" we can make it so that these numbers are exactly matched in magnitude. Then with the correct choice of $\epsilon_i$'s we can make these factors contribute oppositely so that for this particular $F$ and this particular choice of $\epsilon_i$'s the net integral vanishes. 

I think it is always possible to do a similar trick/deformation.

  [1]: https://i.sstatic.net/y1TiQ.jpg