Yes, provided you assume Hermitianity. Then, even more is true.

Take A to be Hermitian (or real symmetric, if you like) matrix. As for B, it can be any positive semidefinite matrix (including your rank 1 case and without regard to the eigenvectors of A). Then your assertion follows from Weyl's Theorem about the eigenvalues of the sum of Hermitian matrices. This is actually stated as Problem 1 on page 198 of the Horn & Johnson *Matrix Theory*.

Here's a Google Books link to it:

http://books.google.ie/books?id=PlYQN0ypTwEC&pg=PA198&dq=interlacing+horn+johnson&hl=en&sa=X&ei=2HZ0T_DrIZOAhQeeqKSmBQ&redir_esc=y#v=onepage&q&f=false

Since B is positive semidefinite, $\lambda_{1}(B) \geq 0$.

As you can see there, you can even bound the $d_{i}$'s from above.