Heavily edited answer, closer to what the question asks.

----

There is an answer to the simpler question:
$$
C (\sigma)^{\gamma} \int_{B_1} u^2 dx  \leq \int_{B_{\sigma}} u^2 dx
$$
where $C$ and $\gamma$ depend only on $\lambda,\Lambda$, and the dimension. Such estimates are available for a general not necessarily symmetric $A$ bounded and measurable in dimension 2 (see [Alessandrini 2010][1]), and for
$A$ Lipschitz continuous in dimension $3$ or more (see Garofalo & Lin 1986 and [Alessandrini, Rondi, Rosset and Vessela][2]). These results are general unique continuation results, not specific to the eigenvalue problem; in that context, the hypothesis on the regularity of $A$ is probably optimal. 
A result slightly closer to what you want is for example
$$
C (\frac{\sigma}{\tau})^{\gamma} \int_{\partial B_{\tau r}} u^2 dx  \leq \int_{\partial B_{\sigma r}} u^2 dx
$$
for $\tau\leq 1/2$, which is proved in the same papers. Both result follow from the the doubling inequality 
$$
C\int_{\partial B_{2\sigma r}} u^2 \leq \int_{\partial B_{\sigma r}} u^2.
$$  
In the other direction (bound from above), you have the three sphere's inequality
$$
\|u\|_{L^2(B_{r_2})} \leq \|u\|^{\alpha}_{L^2(B_{r_1})}\|u\|^{1-\alpha}_{L^2(B_{r_3})}
$$
For every $r_1<r_2<r_3<R$ in $B_R$, with the same assumptions on $A$. But that's not exactly what you wanted, as this is not $\min u$.

Below is an attempt to decide what to expect in terms of exponents in dimension 1. 

What you would like is that the solution behaves reasonably nicely near the boundary $\partial B_r$. 
A counter-example, in dimension 1, would be if the solution was like $u\approx \exp(-\sigma r/n) - \exp(-r/n)$ because in that case the constant $c$ in your estimate would depend in $\sigma$ and $r$ in the wrong way. 

Let us consider the interval $(-1,1)$ (you can always rescale the argument), and the eigenvalue problem $$ - \frac{1}{n^2}(a u')' = \lambda u, \quad u(-1)=u(1)=0,$$
with $1\leq a\leq2$  for example.
By changing variables to $v=u/\sqrt{a}$, you obtain 
$$
- \frac{1}{n^2}v'' + q v = \lambda r v,
$$ 
with $q= \sqrt{a}''/\sqrt{a}$ and $r=1/\sqrt{a}$. 

Now, choose $q,r$ to be discontinuous, and periodic (of period $1$) for $x>0$ and $x<0$. More precisely, choose $q$ to be $q(x)= q_0(nx)$ for $x<0$, and $q(x) = q_0(nx + t)$ for $x>0$ and $r$ of a similar form.

Now, if you try to construct a rescaled 'full space' solution (forgetting the boundary constraint), you can solve it using Floquet Theory, which tells you that the solution will look like 
$$
exp(\theta_{\pm}nx) \psi_{\pm}(nx) \mbox{ for } \pm x >0, \mbox{ with } \psi_{\pm} \mbox{ periodic}.
$$
Playing with the parameter $t$ (try numerically, e.g. with Matthieu's equations for $q_0$ and $r_0$), you can cook-up a case where $\theta_{-}>0$ and $\theta_{+}<0$. Since this 'full space' solution decays exponentially, it is an excellent candidate for the Dirichlet problem, and therefore the real solution will be exponentially close to that, and you are in the bad case previously described. 

It is not a true counter-example, as it uses a very small lower bound for $a$. But I think it tells you that at best with respect to (3) you cannot hope for more than a logarithmic dependence on $(1-\sigma)$ in terms of the  a priori information. 


  [2]: http://arxiv.org/abs/0907.2882

  [1]: http://arxiv.org/pdf/1004.4527v1.pdf