Upper bounds on higher order eigenvalues of regular graphs
I wanted to add this as a comment but it was too long. I am not sure if this completely answers the question. 

For bipartite graphs for instance we can possibly get a simple first cut bound from a simple trace method. Lets look at the adjacency matrix. In your case we want to show that $\lambda_k$ is not close to 0. So consider $Trace(A^2)$. For a d-regular graph this is always $\geq dn$. Therefore $$\sum \lambda_i^2 \geq dn$$ which implies that $$2d^2k + n\lambda_k^2 \geq dn$$ Now for $k &lt;&lt; n$ and $d$ a constant one can essentially get a $\Omega(\sqrt{d})$ bound for $\lambda_k$. So indeed they cant be very close to 1 in the normalized laplacian. I am not sure whether the parameters work out in the way you want them to.