Yes, there is indeed a current in the circuit, however the proposed solution is still valid, even though it requires extra reasoning to justify. I claim that the assumptions $\mathbf J =0$, $\frac{\partial \mathbf E}{\partial t} = 0$ are valid _almost_ everywhere. Specifically, they fail within the conducting contour and in a small vicinity of it, which has the size on the order of the wire diameter. Since we assume wires infinitely thin, macroscopically the assumptions are valid, but it is vital to remember the microscopic details.

I will use Maxwell's equations in Gaussian units, to avoid pesky $\mu_0$'s and $\varepsilon_0$'s. For the reference they look as follows:
$$
\begin{eqnarray}
\nabla \cdot \mathbf E & = & 4\pi \rho \\
\nabla \cdot \mathbf B & = & 0 \\
\nabla \times \mathbf E & = & -\frac{1}{c}\frac{\partial \mathbf B}{\partial t} \\
\nabla \times \mathbf B & = & \frac{4\pi}{c}\mathbf J + \frac{1}{c}\frac{\partial \mathbf E}{\partial t}
\end{eqnarray}
$$

This implies (assuming that $S(t)$ is the region of plane bounded by $L(t)$)
$$
\begin{eqnarray}
\oint_{L(t)} \mathbf B \cdot \mathrm d \mathbf l & = & \iint_{S(t)} (\nabla \times \mathbf B; \mathrm d \mathbf S) \\
& =& \frac{4\pi}{c} \iint_{S(t)} (\mathbf J; \mathrm d \mathbf S) + \frac{1}{c} \iint_{S(t)} \left(\frac{\partial \mathbf E}{\partial t}; \mathrm d \mathbf S \right)
\end{eqnarray}
$$

However all current flows in the plane $S(t)$, thus its flux through $S(t)$ is $0$.

The term with the partial derivatives is harder to study. First note that _macroscopically_ there are no free charges, in the sense that the macroscopic charge distribution is constant in time. Also the system is quasi-stationary since the speed $v \ll c$ --- this allows us to exclude any EM waves from the problem and only work with charges and currents. This implies that macroscopically $\mathbf E$ is stationary, but if we would assume globally $\frac{\partial \mathbf E}{\partial t} = 0$, then Maxwell's equations would imply that $\mathbf B$ and $\mathbf J$ are also stationary and always $0$. To see that this isn't true we need to consider what happens in the wire itself.

We assume the wire to be an ideal conductor with zero resistance. Ohm's law says that in the wire $\mathbf E = \rho \mathbf J$, if $\rho = 0$ then finite current implies _$\mathbf E = 0$ within the wire_ (EDIT: since we are only interested in the vertical component of $\mathbf E$ and there can be no vertical current, $E_z = 0$ in wire even if $\rho \ne 0$). Thus we see that even before the movement starts the field isn't equal to $\mathbf E_0$ everywhere --- it is $0$ within the wire and has some intermediary value in its vicinity. This also shows that globally $\mathbf E$ isn't stationary --- the movement of the wire causes the movement of the zeroes of $\mathbf E$ and of the shielding charges in the wire. This, in turn, causes the magnetic field and the induced current. If we try to calculate the derivative, then we see that near the wire $E$ changes from $E_0$ to $0$ over an infinitely small interval of time, so the derivative has some [delta-function](https://en.wikipedia.org/wiki/Dirac_delta_function)-like form, providing some finite (generally) non-zero value to the integrals.

To calculate the surface integral of $\frac{\partial \mathbf E}{\partial t}$, we need to convert it to a more manageable form, something like a derivative of a continuous function. The general formula for a full derivative of a time-dependent surface integral is 
$$
\frac{\mathrm d}{\mathrm d t} \iint_{S(t)} (\mathbf F ; \mathrm d \mathbf S)
= \iint_{S(t)} \left( \frac{\partial \mathbf F}{\partial t}; \mathrm d \mathbf S \right ) + \frac{1}{\mathrm d t}\iint_{\delta S(t)} (\mathbf F; \mathrm d \mathbf S)
$$

Here $\delta S(t)$ is the infinitesimal variation of the surface $S(t)$ and I assume that $S(t)$ varies by adding extra area, like in the problem (i.e. no movement of the interior). This is just the usual product rule for the calculation of derivatives. In our problem $\mathbf F = \mathbf E$ and the surface is chosen so that its boundary passes inside the wire loop. This means that $\mathbf E =0 $ near the boundary of $S(t)$ and thus the integral over the variation of area is $0$, so the second term vanishes and we have
$$ 
\iint_{S(t)} \left( \frac{\partial \mathbf E}{\partial t}; \mathrm d \mathbf S \right )  
= 
\frac{\mathrm d}{\mathrm d t} \iint_{S(t)} (\mathbf E ; \mathrm d \mathbf S)
$$

Since $\mathbf E$ is everywhere bounded and almost everywhere equal to $\mathbf E_0$, the answer to your problem follows.

Note that the circulation of $\mathbf B$ doesn't depend on the specific contour passing through the wire, but will change if we move the contour outside the wire. Also note that if we consider an infinitely small loop around a section of the wire, then the circulation of $\mathbf B$ will be finite nonzero if there is a non-zero current passing through the section. This demonstrates that when the problem asks an integral around the perimeter, we must consider _exactly_ the perimeter, even a small variation would give an incorrect answer.