First, consider the positive and negative charges in your moving wire. Since they are moving in a (obviously non-conservative) magnetic field, they experience Lorentz's force $q \ \mathbf{v} \times \mathbf{B}$ which is, in your picture, upwards for positive (and downwards for negative) charges. So they will be accelerated in exactly the same way (for whatever movement your wire gets) as if they were experiencing the electric field that you could calculate using the flux integral variation.
On the other hand, by changing to a moving reference frame, you transform any magnetic field into an electric field - and vice versa (Lorentz transformation). So, in a frame moving with the wire you see the magnetic field as a non-conservative electric field, and this E field accelerates your charges. That's what creates the current in your circuit.
Of course, after a short transient phase where your charges accelerate, you get (because of collisions) a constant current - that's basic Ohm's law here.
And the important point is, whatever your point of view, you will always find the exact same motion for the charges.
Now, neither the magnetic field nor the electric field that appears in the moving frame are conservative (the latter does not appear from Coulomb's law, which in this case states $\nabla \cdot E=0$, but from induction)
The magnetic field is perpendicular to the current's direction, and the current is going normal to the loop, meaning that $B$ is parallel to the loop's plane. Because flux is the dot product of the magnetic field and the loop's normal "surface vector", and the angle is $90^\circ$, it is equal to $0$, thus not changing in time and not inducing an emf.
$$ \varepsilon = -\frac{d \Phi}{dt} = -\frac{d}{dt} (0) = 0 $$
Best Answer
The magnetic field actually is in the same direction on each side of the plane. You can see this in the image below and as explained here. As you can see, there is definitely a magnetic flux through the area bound by the coil itself.
Yes, you are right. Gauss's law for magnetic fields tells us $$\oint\mathbf B\cdot\text d\mathbf A=0$$
but this is a surface integral over a closed surface. The area bound by the coil is not a closed surface, so we don't need to worry about this applying here.
So, even though the flux through thiis area is in fact not $0$, I will address a concern you seem to have in linking these two ideas together. You seem to be thinking that a $0$ flux means that the surface integral must have been done over a closed surface. This is not the case. The statement "If the integral is over a closed surface, then the magnetic flux is $0$" is not a biconditional statement. In other words, the statement "If the magnetic flux is $0$ then the integral was done over a closed surface" is a false statement.