On the determinant of a certain matrix with non negative integer entries with fixed row sum

Yes, it's true. By assumption, $A$ has nonzero determinant so has an inverse $A^{-1}$. Then \begin{equation} \det(A+B)=\det(A)\det(I+A^{-1}B)=d\det(I+A^{-1}B). \end{equation} To prove the claim, it suffices to show that this latter determinant is $1+m/d$.

Think about what $A^{-1}B$ looks like; it only has nonzero entries on the first column. So $I+A^{-1}B$ is lower triangular, and the diagonal is all $1$s except for the $(1,1)$ entry, where it is $1$ plus the first entry in $m\cdot A^{-1}e$, where $e$ is the all-ones vector (as this is the first column of $B$ by assumption). Because the determinant of a lower triangular matrix is the product of the diagonal entries, it suffices to show that $A^{-1}e=(1/d)e$. But this is true as $e$ is an eigenvector of eigenvalue $d$ for $A$, as $A$ has constant row sum $d$. Thus, $A^{-1}e=(1/d)e$, which completes the proof.

(By the way, this evidently also means that the $n\geq 3$, non-negativity and integrality of $A$, and positivity and integrality of $B$ are not necessary. Actually, even better, if you allow $A$ to have arbitrary nonzero determinant $c$, not necessarily equal to the common row sum $d$, you will get $\det(A+B)=c(1+m/d)$.)