It is not true that "potential difference makes no sense when the curl of $\vec E$ is nonzero". Instead, the scalar and vector potentials $\Phi,\vec A$ may always be defined and the electric field may be expressed in terms of these variables as
$$ E = -\nabla \phi - \frac{\partial A}{\partial t} $$
In fact, these potentials aren't unique – any other choice of the potentials related to the original one by a gauge transformation is equally good to calculate the same $\vec E$ and $\vec B = \text{curl }\vec A$.
In an rlc circuit, one may always confine the nonzero $\vec A$ to a small vicinity of the inductor which also determines the right gauge transformation. Then the potential difference across any element of the circuit is well-defined and the sum of these differences over a closed loop is zero.
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)
Best Answer
Short answer:
No, you can't say that. "Conservative" is an attribute of the electric field and not of its line integrals.
Long answer:
It is true that some electric fields are conservative while others are not. In particular, all electrostatic fields (that is, fields generated by non-accelerating charges) are conservative. A conservative field is such that its line integral on a closed path is always zero:
$$ \oint \vec{E} \cdot d\vec{s} = 0. $$
This relation can be also written in a local form by means of Stokes' theorem:
$$ \vec{\nabla} \times \vec{E} = 0. $$
A consequence of being conservative is that it is possible to define a potential difference between two points $A$ and $B$ which is independent of the path chosen to connect them. In other words, the quantity
$$ \Delta V_{AB} \equiv \int_A^B \vec{E} \cdot d\vec{s} $$
can be written as the difference between a function evaluated in $A$ and $B$, that is
$$ \Delta V_{AB} = V(A) - V(B). $$
Here is the important bit: the line integral can always be computed, even when the electric field is not conservative. However, in this case the final result depends on the path over which the line integral is computed, and hence no function $V$ (the electrostatic potential) can be defined.
The electromotive force, on the other hand, is defined as
$$ f_{\rm em} = \oint \vec{E} \cdot d\vec{s}, $$
which, by definition, can be different from zero only if $\vec{E}$ is not conservative. However, note that the closed path integral above might be zero for some paths even when $\vec{E}$ is not conservative.