By [Gerschgorin's Circle Theorem][1], if the number of nonzero entries in each row is at most $d$, the spectral radius is at most $d$.  This bound is attained e.g. in a case where there are $d$ $1$'s  in each row and no $-1$'s, with eigenvector $(1,\ldots,1)^T$ for eigenvalue $d$.

EDIT: If the matrix has at most $d$ nonzero entries, the spectral radius is at most $ (d+1)/2$.  This follows from the following facts:

 1. The spectrum is contained in the [numerical range][2], so the spectral radius is at most the numerical radius.
 2. The numerical radius is a norm.
 3. The numerical radius of a diagonal matrix with entries of $\pm 1$ and $0$ on the diagonal is $1$.
 4. The numerical radius of a matrix with a single off-diagonal entry of $\pm 1$ and all else $0$ is $1/2$.

This estimate does not seem to be sharp (except of course if $d=1$).

 


  [1]: http://en.wikipedia.org/wiki/Gershgorin_circle_theorem
  [2]: http://en.wikipedia.org/wiki/Numerical_range