This is not an answer to the question, just an explicit calculation to see what shows up in a discontinuous case.
Let us look at a one dimensional piecewise constant case, namely the domain is $(-r,r)$, and the coefficient $A(x)=a$ when $x<0$, and $A(x)=b$ when $x>0$.
Let us now consider the first eigenvalue/eigenvector of the problem $$ -\frac{d}{dx}(A(x)\frac{d}{dx} u) = \frac{\lambda}{r^2} u \mbox{ in }H^1_0(-r,r), $$ normalised by $u(0)=1$. A simple comparison using the Raleigh quotient shows that $\lambda=\frac{\pi^2}{4}\nu$ with $$ \min(a,b)\leq \nu \leq \max(a,b) $$
An explicit computation shows that $u,\nu$ are given by $$ u=\frac{\displaystyle\cos\left(\frac{\pi\sqrt{\nu}}{2\sqrt{a}}\right)\sin\left(\frac{\pi\sqrt{\nu}x}{2\sqrt{a}r}\right)}{\displaystyle\sin\left(\frac{\pi\sqrt{\nu}}{2\sqrt{a}}\right)}+\cos\left(\frac{\pi\sqrt{\nu}x}{2\sqrt{a}r}\right)\mbox{ when }x<0, $$ and $$ u=-\frac{\displaystyle\cos\left(\frac{\pi\sqrt{\nu}}{2\sqrt{b}}\right)\sin\left(\frac{\pi\sqrt{\nu}x}{2\sqrt{b}r}\right)}{\displaystyle\sin\left(\frac{\pi\sqrt{\nu}}{2\sqrt{b}}\right)}+\cos\left(\frac{\pi\sqrt{\nu}x}{2\sqrt{b}r}\right)\mbox{ when }x>0, $$ and $\nu$ is the smallest positive solution of $$ \tan\left(\frac{\pi\sqrt{\nu}}{2\sqrt{b}}\right)\sqrt{\nu}\sqrt{a}=-\tan\left(\frac{\pi\sqrt{\nu}}{2\sqrt{a}}\right)\sqrt{\nu}\sqrt{b}. $$ It is easy to see that for the solution to stay positive, it must stay fairly close to the $\min(a,b)$. For example, when $b/a \gg1$, $\nu\approx 4a$, as the solution concentrates in the half domain where the conductivity is the lowest, and becomes close to $\sin(\frac{\pi}{R}(x+R))$. Near the boundaries: we have at $-r$ $$ u\approx \frac{\cos^2\left(\frac{\pi\sqrt{\nu}}{2\sqrt{a}}\right)}{\sin\left(\frac{\pi\sqrt{\nu}}{2\sqrt{a}}\right)}(1+\frac{x}{r}) $$ and at $r$ $$ u\approx \frac{\cos^2\left(\frac{\pi\sqrt{\nu}}{2\sqrt{b}}\right)}{\sin\left(\frac{\pi\sqrt{\nu}}{2\sqrt{b}}\right)}(1-\frac{x}{r}) $$
(edited) So the scaling in $r$ of the dependence is still the one you hope for.