Here is a first-order approximation.
I'll write $\tilde{v} = v + \rho w + O(\rho^2)$ and $\tilde{\lambda} = \lambda + \rho \mu + O(\rho^2)$.  We may assume $\|v\| = \|\tilde{v}\| = 1$, so $v^T w = 0$.   If $\lambda$ is a simple eigenvalue of $B$, $B - \lambda I$ is invertible on the orthogonal complement $v^\perp$ of $v$.  Taking $(B + \rho b b^T) \tilde{v} = \tilde{\lambda} \tilde{v}$ to first order in $\rho$, we have
$$ B w + b b^T v = \mu v + \lambda w $$
Let $c = b - (b^T v) v$ which is orthogonal to $v$, so that this equation splits into
$$ B w + (b^T v) c = \lambda w \ \text{and} (b^T v)^2 v = \mu v$$
Thus $\mu = (b^T v)^2$ and $w = - (b^T v) (B - \lambda I)|_{v^\perp}^{-1} c$.