Let $\lambda, \mu$ be the Perron-Frobenius eigenvalues of two non-negative matrices $A,B$ respectively. I am interested in knowing whether there are any results available on the upper bound of $|\lambda-\mu|$ in terms of norms of $\|A-B\|$.



upper bound on the difference between two Perron-Frobenius eigenvalues
I guess you want a bound in terms of a norm of $A-B$, since one obviously has $|\rho(B) - \rho(A)| \leq ||A|| + ||B||$ for any operator norm $|| \cdot ||$. 


If $A$ is irreducible, then its Perron eigenvector $x\_A$ (normalized so that $||x\_A||\_1 = 1$) has strictly positive entries, hence $K(A) = \max\_i (x\_A)\_i^{-1} < \infty$. In that case one has
$$
(\*) \ \ \ \ \ \ \ \ \|\rho(B) - \rho(A)| \leq K(A) ||B - A||\_{\infty}, 
$$
where $||C||\_{\infty} = \max\_{i,j} |c\_{ij}|$. Indeed,
$$
(\rho(B) - \rho(A)) \langle x\_{A} | x\_{B^\*} \rangle = \langle x\_{A} | B^\* x\_{B^\*} \rangle - \langle A x\_{A} | x\_{B^\*} \rangle = \langle (B-A) x\_{A} | x\_{B^\*} \rangle,
$$
and the conclusion follows from the inequalities 
$$
\langle x\_{A} | x\_{B^\*} \rangle \geq K(A)^{-1} ||x\_{B^\*}||\_1 = K(A)^{-1}
$$
and
$$
|\langle (B-A) x\_{A} | x\_{B^\*} \rangle | \leq || B-A||\_{\infty} ||x\_A||\_1 ||x\_{B^\*}||\_1 = || B-A||\_{\infty} .
$$


Note that $(\*)$ is still valid for general $A$, if one replaces $K(A)$ by $\max(K(A),K(B)$, with $K(A)= \max\_{j} K(A\_j)$, where $(A\_j)\_j$ are the irreducible blocks of $A$.


