Yes. Take two $a,b$ generators of $S_n$, where $b$ has odd order and fixes point $1$. For example, if $n$ is even, let $a=(1,2)$, $b=(2,3,\ldots,n)$. If $n$ is odd let $a=(1,2,3,4)$, $b=(3,4,\ldots,n)$.

Let $\bar{a},\bar{b}$ be their natural images$^\dagger$ in $B_n$. Then $B_n = \langle \bar{a}, \bar{b}(1,-1) \rangle$. This is because $\bar{b}$ and $(-1,1)$ commiute and have coprime order, so both $\bar{b}$ and $(-1,1)$ are powers of $\bar{b}(-1,1)$ and hence $\langle \bar{a}, \bar{b}(1,-1) \rangle=\langle \bar{a}, \bar{b},(1,-1) \rangle =B_n$.

$\dagger$ : For a permutation $\alpha$ of $\{1,2,\ldots,n\}$ let $\alpha'$ be the permutation of $\{-1,-2,\ldots,-n\}$ defined by $\alpha(-i) := -\alpha(i)$ for $1 \le i \le n$. By the natural embeddding $S_n \to B_n$ I mean the map defined by $\alpha \mapsto \alpha\alpha'$.