I strongly suspect the answer is yes. By the min-max theorem, the largest eigenvalue of a hermitian matrix $M$ is $$ \lambda_{max}=\text{max} \left( \frac{x^*Mx}{x^*x} \right) $$ This is also its spectral radius. It seems intuitively that $\lambda_{max}$ could not decrease if some entries of $M$ increased in magnitude (in such a way that $M$ remains hermitian of course), but I cannot prove it.
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This is not true even for positive-definite real matrices. If $\lambda$ is the leading eigenvalue with eigenvector $v$ (and assuming that this eigenvalue is simple) then the derivative of $\lambda$ with respect to the $(i,j)$th entry of $M$ is given by $$v^T(e_{i} \otimes e_{j}) v = v_iv_j.$$
Of course to keep $M$ symmetric you need to vary both the $(i,j)$ and $(j,i)$th entry and will get a factor of $2$, but in any case generally these derivative will not all have the correct sign. As a concrete example consider $$\left[\begin{array}{ccc}3 & -1 & 1\\-1 & 2 & 1\\1 & 1 & 2\end{array}\right].$$ Increasing the $(2,3)$ and $(3,2)$ entries decreases the spectral radius.
One case where your property will hold is if the matrix has all positive entries, since the leading eigenvector will have all positive entries by the Perron–Frobenius theorem.
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