This answer addresses part of the question. It doesn't address Wick rotation, except for brief comments at the end.
Part of the question is why the quantity $\langle\varphi_1|U(0,t)|\varphi_2\rangle$ is independent of how the spacetime between times $0$ and $t$ is foliated by spacelike slices. It must be, because this quantity can be written as a path integral (as shown in the OP) which is more clearly independent of the foliation.
But suppose we didn't know that it could be written as a path integral. How can we see more directly that $\langle\varphi_1|U(0,t)|\varphi_2\rangle$ is independent of the foliation? In the usual foliation, the Hamiltonian $H$ is independent of time, and we have $U(0,t)=\exp(-iHt)$. However, $H$ depends on the foliation, and for a more general foliation, $H$ is not independent of the time coordinate. So why is $U(0,t)$ be independent of the foliation?
Intuition (Lorentzian case)
Here's some intuition. The Hamiltonian can be expressed in terms of the stress-energy tensor $T^{\mu\nu}(x)$. The components of the stress-energy tensor generate "local translations" in spacetime, and mutually commuting components can be applied in any order we want. The mutually commuting condition is ensured by spacelike separation, so we can apply "local time translations" in spacelike-separated regions in any order we want. This corresponds to the freedom to change the foliation by spacelike hypersurfaces. The sequence of slices must be traversed sequentially so that we don't violate the mutually-commuting constraint, but we can deform the slices without changing the answer as long as they remain spacelike (and as long as the initial and final slices don't change, of course).
Explicit formulation (Lorentzian case)
In case the intuition isn't convincing, here's a more explicit version. It's still not rigorous, but maybe a real mathematician could turn this loose outline into a legitimate proof.
For any foliation, we can write the corresponding $H$ in terms of $T^{\mu\nu}$ like this:
$$
\newcommand{\la}{\langle}
\newcommand{\ra}{\rangle}
\newcommand{\pl}{\partial}
H(x^0) = \int_\Sigma\ \sqrt{|\det{g}|}\,T^{0\mu}n_\mu
\tag{1}
$$
where $\Sigma$ is a spacelike hypersurface (one of the slices in the foliation), $g$ is the metric tensor restricted to $\Sigma$, and $n_\mu$ is the normal to $\Sigma$. Both $T$ and $n_\mu$ are functions of the coordinates on $\Sigma$. The coordinate system is such that different values of $x^0$ give different slices $\Sigma$ in the foliation. The metric tensor associated with this coordinate system is just a coordinate transform of the original Minkowski metric. The factor $\sqrt{|\det{g}|}$ makes the integral invariant under coordinate transformations on $\Sigma$ that preserve $x^0$. In general, the resulting $H$ is a function of the time-coordinate $x^0$ in addition to depending on the foliation.
Since $H$ depends on $x^0$, we can no longer express $U(0,t)$ as a simple exponential of $H$, but we can define it by these conditions:
$$
\frac{d}{dx^0}U(0,x^0)=-iH(x^0)U(0,x^0)
\hskip2cm
U(0,0) = 1.
\tag{2}
$$
This implies
$$
U(0,t) = U(0,x^0)U(x^0,\,x^0+dx^0)U(x^0+dx^0,\,t).
\tag{3}
$$
That's useful because we can consider a foliation-change that only affects the middle factor, and we can take the interval $dx^0$ to be arbitrarily short, because any foliation can be reached from any other through a sequence of such small deformations. For infinitesimal $dx^0$, we have
\begin{align}
U(x^0,\,x^0+dx^0)
&\approx 1 - iH(x_0)dx^0 \\
&= 1 - i\,dx^0\int_\Sigma\ \sqrt{|\det{g}|}\,T^{0\mu}n_\mu.
\tag{4}
\end{align}
The change in (4) induced by an infinitesimal change in $\Sigma$ is
\begin{align}
\delta U(x^0,\,x^0+dx^0)
&\approx - i\,dx^0\oint_{\pl M}\ \sqrt{|\det{g}|}\,T^{0\mu}n_\mu \\
&\approx - i\,dx^0\int_M\ \pl_\mu\big( \sqrt{|\det{g}|}\,T^{0\mu}\big)
\tag{5}
\end{align}
where $\pl M$ is the closed hypersurface formed by $\Sigma$ and its infinitesimally deformed version and where $M$ is the part of spacetime bounded by $\pl M$. Now use the identity
$$
\pl_\mu\big( \sqrt{|\det{g}|}\,T^{0\mu}\big)
=
\sqrt{|\det{g}|}\,\nabla_\mu\,T^{0\mu}
\tag{6}
$$
where $\nabla$ is the Levi-Civita connection associated with $g$. (This is equation (3.4.10) in Wald's General Relativity.) This is zero because of the local conservation law $\nabla_\mu T^{\mu\nu}=0$, which is just the coordinate transform of the usual conservation law $\partial_\mu T^{\mu\nu}=0$ from Minkowski spacetime. So (5) is zero.
Altogether, this indicates that $U(0,t)$ is independent of how the spacetime between $0$ and $t$ is foliated by spacelike hypersurfaces. (I said "indicates" because this wasn't a rigorous proof, just an outline.)
Wick rotation
After Wick rotation, we no longer have a distinction between timelike and spacelike in the path integral. What happens to $\langle\varphi_1|U(0,t)|\varphi_2\rangle$ in this case?
Recall that in the operator formulation in the Lorentzian case, we have two distinct notions of ordering: operator ordering, and time ordering. In the path integral, these two orderings are equated: the path integral formulation of a correlation function automatically corresponds to a time-ordered correlation function in the operator formulation. An interesting exercise is to think about how the canonical commutation relations are encoded in the path-integral formulation (hint: use explicit infinitesimal time-differences to dictate the order of the operators in the operator products) and then to think about what happens to that in the Wick-rotated case. I won't try to address that here, partly because this post is already long and partly because I'd need to think through all the implications of Wick rotation more carefully myself, but I'm pretty sure this would be an enlightening exercise.
