= Solution
Write $s=A(S)=4\pi$, $L=L(\partial U)$, and $a=\min\{A(U),A(S\setminus U)\}$. We first prove a constant-version <spherical isoperimetric inequality> directly, including regions with several boundary components.
If $L\ge1$, the arithmetic bound $A(U)A(S\setminus U)\le s^2/4=4\pi^2$ already gives $A(U)A(S\setminus U)\le4\pi^2L^2$. If $L=0$, a region with smooth boundary is empty or the whole connected sphere up to boundary-null changes, and its area product is zero. For $0<L<1$, write the smooth boundary as finitely many disjoint <simple closed curves> $\gamma_j$ of lengths $l_j$. Fix a point $p_j$ on each curve. The shorter boundary arc from $p_j$ to any other point has length at most $l_j/2$, so $\gamma_j$ is contained in the closed geodesic ball of radius $l_j/2$ about $p_j$.
The complement of that small ball is connected and lies entirely on one side of $\gamma_j$. The <Jordan curve theorem> therefore gives a disk $E_j$ bounded by $\gamma_j$ contained in the ball. Its area satisfies
$$
A(E_j)\le2\pi(1-\cos(l_j/2))\le\frac\pi4l_j^2.
$$
The <indicator function> of $U$ and the sum $\sum_j\mathbf1_{E_j}$ have the same jumps modulo two across every boundary curve. Their difference modulo two is consequently constant throughout the sphere off the boundary. Thus either $U$ or its complement lies in $\bigcup_jE_j$, up to its area-zero boundary. This also handles nested curves and disconnected regions. It follows that
$$
a\le\sum_jA(E_j)\le\frac\pi4\sum_jl_j^2\le\frac\pi4L^2,\qquad A(U)A(S\setminus U)\le sa\le\pi^2L^2.
$$
Combining the cases, \b[one valid universal choice is $\kappa=4\pi^2$]. Sharpness is not required.
For the inclusion estimate put $x=A(V)$, $d=A(D)>0$ and $y=A(V\cap D)$. Splitting $D$ into its portions inside and outside $V$ gives
$$
xd-sy=xA(D\setminus V)-(s-x)A(D\cap V).
$$
Because $0\le A(D\setminus V)\le s-x$ and $0\le A(D\cap V)\le x$, this expression lies between $-x(s-x)$ and $x(s-x)$. Apply the preceding <spherical isoperimetric inequality> to $V$ and divide by $sd$:
$$
\boxed{\left|\frac{A(V)}{A(S)}-\frac{A(V\cap D)}{A(D)}\right|\le\frac{\kappa L(\partial V)^2}{A(S)A(D)}.}
$$
Thus the constant in the <spherical inclusion area discrepancy> is independent of the nonempty open set $D$.
Let $g$ and $\widetilde g$ be the old and new smooth <Riemannian metrics>. Their area forms satisfy
$$
d\widetilde A=\theta\,dA,\qquad \theta=\sqrt{\frac{\det\widetilde g}{\det g}}>0.
$$
The determinant ratio is invariant under changes of coordinates, so $\theta$ is a globally defined smooth <function>. The <layer cake representation> follows from $\theta(x)=\int_0^\infty\mathbf1_{\{\theta(x)>t\}}\,dt$ and <Tonelli's theorem>:
$$
\boxed{\widetilde A(U)=\int_U\theta\,dA=\int_0^\infty A(U\cap D_t)\,dt,\qquad D_t=\{x:\theta(x)>t\}.}
$$
Smoothness makes every $D_t$ open. Compactness of the sphere gives $0<m=\min\theta\le M=\max\theta<\infty$. Multiply the inclusion bound by $A(D_t)$ before integrating; for empty $D_t$ the resulting bound is trivially zero. For $t<m$, $D_t=S$ and the discrepancy is exactly zero; for $t>M$, $D_t$ is empty. Therefore
$$
\left|\frac{x}{s}\widetilde A(S)-\widetilde A(V)\right|\le\int_m^M\left|\frac{x}{s}A(D_t)-A(V\cap D_t)\right|dt\le\frac{\kappa(M-m)}s L(\partial V)^2.
$$
Dividing by $\widetilde A(S)>0$ proves
$$
\boxed{\left|\frac{A(V)}{A(S)}-\frac{\widetilde A(V)}{\widetilde A(S)}\right|\le\frac{C'L(\partial V)^2}{A(S)\widetilde A(S)},\qquad C'=\kappa(M-m).}
$$
This constant depends only on the two metrics. If $\theta$ is constant the normalized areas agree exactly, as the bound also shows.
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