This is not a complete answer. It's more of a, hopefully, helpful observation:
I'm not convinced that for a loop that is not fixed in time, the equation you wrote down is correct. Here's why.
Edit 1. I recant my position on this, please read on and see Edit 2 below!
The differential form of Faraday's law is
$$
\nabla\times\mathbf E = -\frac{\partial\mathbf B}{\partial t}
$$
Now let $C_t$ denote a closed curve (topologically a circle) at any given time $t$, and let $\Sigma_t$ be any surface (topologically a disk) whose boundary is $C_t$. Then integrating both sides over $\Sigma_t$ and using Stokes' theorem gives
$$
\int_{C_t} \mathbf E\cdot d\boldsymbol\ell = -\int_{\Sigma_t}\frac{\partial\mathbf B}{\partial t}\cdot d\mathbf a
$$
The expression you wrote down follows from taking the time derivative outside of the integral and noting that what remains is the flux. However, if the surface over which the flux is being computed is changing, then you can't take the derivative outside of the integral without incurring an extra term. In fact, one has
$$
\frac{d}{dt}\int_{\Sigma_t}\mathbf B\cdot d\mathbf a = \int_{\Sigma_t}\frac{\partial \mathbf B}{\partial t}\cdot d\mathbf a + \eta(t)
$$
where the expression for $\eta$ is given in the Appendix below. It follows from this that for a time-varying loop, Faraday's law is
$$
\int_{C_t} \mathbf E\cdot d\boldsymbol\ell = -\frac{d}{dt}\int_{\Sigma_t}\mathbf B\cdot d\mathbf a + \eta(t)
$$
If the loop lies in a plane, and if $\mathbf B$ points perpendicular to the surface $\Sigma_t$ lying in the plane whose boundary is the loop, and if the magnetic field is spatially constant on this surface, then this reduces to
$$
\int_{C_t} \mathbf E\cdot d\boldsymbol\ell = -\frac{d}{dt}\Big(B(t)A(t)\Big) + \eta(t)
$$
and it seems to me like your expression is therefore missing the extra $\eta(t)$ term.
Edit 2. Prompted by Art Brown's comments, upon examining the Leibniz integral formula it becomes clear that the expression for $\eta(t)$ can be written in a vastly simpler (and manifestly gauge-invariant which is nice) form than in my appendix;
$$
\eta(t) = -\int_{C_t} \mathbf v\times\mathbf B\cdot d\boldsymbol\ell, \qquad \mathbf v = \frac{\partial\gamma}{\partial t}
$$
If you'd like to see the proof of this, take a look at the question I asked on math.SE when I got confused about the proof: https://math.stackexchange.com/questions/453297/leibniz-integral-rule-confusion/453316?noredirect=1#453316
What's interesting about this is that the integral form of Faraday's Law can then be written as follows:
$$
\int_{C_t}(\mathbf E +\mathbf v\times\mathbf B)\cdot d\boldsymbol\ell = -\frac{d}{dt}\int_{\Sigma_t} \mathbf B\cdot d\mathbf a
$$
but notice that the left-hand integrand is simply the Lorentz force per unit charge, and it is precisely this quantity that when integrated around the loop at a given time gives the emf $\mathcal E$, the emf is not just the line integral of the electric field around the loop, this is only true when the loop is stationary! The punchline is therefore that the expression
$$
\mathcal E=-\frac{d\Phi}{dt}
$$
is actually valid for a loop with arbitrary movement and deformation, you just have to be careful about the definition of EMF.
Appendix.
Let $\boldsymbol\gamma(t,\lambda)$ denote a function that for each $t$ traverses the curve $C_t$ once as $\lambda$ goes from parameter value $0$ to $1$, then
$$
\eta(t) = \int_0^1d\lambda \left(\frac{\partial\gamma_i}{\partial \lambda}(t,\lambda)\frac{\partial\boldsymbol\gamma}{\partial t}(t,\lambda)\cdot \nabla A_i(t,\boldsymbol\gamma(t,\lambda) + A_i(t,\boldsymbol\gamma(t,\lambda)\frac{\partial^2\gamma_i}{\partial \lambda\partial t}(t,\lambda)\right)
$$
where $A_i$ are the components of the magnetic vector potential. We derive this expression as follows. First, recall that the magnetic field can be written as the curl of a vector potential;
$$
\mathbf B = \nabla\times\mathbf A
$$
This tells us that
\begin{align}
\int_{\Sigma_t}\mathbf B\cdot d\mathbf a
&= \int_{\Sigma_t}\nabla\times\mathbf A\cdot d\mathbf a
= \int_{C_t}\mathbf A\cdot d\boldsymbol \ell
= \int_0^1 A_i(t,\boldsymbol\gamma(t,\lambda))\cdot\frac{\partial\gamma_i}{\partial\lambda}(t,\lambda)d\lambda
\end{align}
where is the last equality I restored some suppressed arguments of functions for clarity in the next step, and I rewrote the dot product in components. Now we take the time derivative of both sides. We need to use the product rule and the chain rule, and this results in the following expression:
\begin{align}
\frac{d}{dt}\int_{\Sigma_t}\mathbf B\cdot d\mathbf a
&= \int_0^1 \left(\frac{\partial A_i}{\partial t}\cdot\frac{\partial\gamma_i}{\partial\lambda}+\frac{\partial\boldsymbol\gamma}{\partial t}\cdot \nabla A_i\frac{\partial\gamma_i}{\partial \lambda}+A_i\frac{\partial^2\gamma_i}{\partial t\partial\lambda}\right)d\lambda\\
&= \int_0^1 \frac{\partial A_i}{\partial t}\cdot\frac{\partial\gamma_i}{\partial\lambda}d\lambda + \int_0^1 \left(\frac{\partial\boldsymbol\gamma}{\partial t}\cdot \nabla A_i\frac{\partial\gamma_i}{\partial \lambda}+A_i\frac{\partial^2\gamma_i}{\partial t\partial\lambda}\right)d\lambda \\
&=\int_{C_t} \frac{\partial\mathbf A}{\partial t}\cdot d\boldsymbol\ell + \underbrace{\int_0^1 \left(\frac{\partial\boldsymbol\gamma}{\partial t}\cdot \nabla A_i\frac{\partial\gamma_i}{\partial \lambda}+A_i\frac{\partial^2\gamma_i}{\partial t\partial\lambda}\right)d\lambda}_{\equiv \eta(t)}\\
&= \int_{\Sigma_t}\frac{\partial\mathbf B}{\partial t}\cdot d\mathbf a + \eta(t)
\end{align}
