Do the integral forms of Maxwell's Equations have limited applicability because of retardation?

It is easy to show that the differential and integral forms of Maxwell's equations are equivalent using Gauss's and Stokes's theorems.

Correct, they are equivalent (assume no GR, and no QM) in the sense that if the integral versions hold for any surface/loop then the differential versions hold for any point, and if the differential versions hold for every point then the integral versions hold for any surface/loop. (This also assumes you write the integral versions in the complete and correct form with the flux of the time partials of the fields and/or with stationary loops.)

Suppose there is a time-varying current in a wire $I(t)$ and I wish to find the fields a long way from the wire.

Ampere's Law is correct, but it is not also helpful. If you know the circulation of $\vec B$, you can use it to find the total current (charge and displacement). If you know the total current (charge and displacement), then you can find the circulation of $\vec B$. But solving for $\vec B$ itself is hard unless you have symmetry.

What about using Ampere's law in integral form? What is the limit of its validity?

It is completely valid, but it might not be helpful. When you write: $$ \oint \vec{B}(r,t)\cdot d\vec{l} = \mu_0 I(t) + \mu_0 \iint \epsilon_0 \frac{\partial \vec{E}(r,t)}{\partial t} \cdot d\vec{a}$$ then the $t$ that is used on each side of the equation is exactly the same.

When $I(t)$ changes, then $\vec B$ field nearby changes quickly, and when there is a changing $\vec B$ field there is a circulating electric field, so as the region of changing $\vec B$ field expands, so does the region of newly circulating electric fields. Both expand together. Eventually the expanding sphere of changing $\vec B$ field and changing circulating electric fields finally starts to reach the Amperian loop (together), and only then does the circulation of $\vec B$ on the far away Amperian loop change. If there was just one change in $I(t)$, then the expanding shell of changing electric fields continues expanding, and you are stuck with the new value of the circulating $\vec B$ field, based on the current that changed a while back.

So, to solve for $\vec B$, you'd need both $I$ and $\partial \vec E /\partial t$ and the latter you need for all the empty space on a surface through the Amperian Loop. Maxwell's equations don't have limited validity and do not need to be modified. They just aren't always as useful as you might want them to be.

I don't really see what the problem is. If $B$ is zero on the boundary, your equation shows $0= \mu_0 I(t) + \mu_0 \iint \epsilon_0 \frac{\partial \vec{E}(r,t)}{\partial t} \cdot d\vec{A}$, so $$\iint \epsilon_0 \frac{\partial \vec{E}(r,t)}{\partial t} \cdot d\vec{A}=-I(t)$$

Why is it that you expect that this equation isn't satisfied? I can tell very little about the quantity $\frac{\partial \vec{E}(r,t)}{\partial t}$

The differential versions of the Maxwell equations imply the integral versions, and the integral versions imply the differential versions, so you can't break one without breaking the other.

Not all systems of PDEs are local wave equations, so the fact that Maxwell's equations can be phrased in terms of PDEs (local laws) doesn't mean that effects are local. Conversely, the fact that Maxwell's equations can be phrased in terms of integrals ("global" laws) doesn't mean that the effects are global.