The answer is no even for functions with values in $L_\infty(\mu)$ (for the purpose of this question embedded in $B(H)$), as long as the cardinality of the space $L_\infty(\mu)$ is larger than continuum. The following argument works for many reasonable topologies, including the w.o.t. in the question.

Let $\Omega$ be the set $L_\infty(\mu)\times L_\infty(\mu)$ with the smallest $\sigma$-algebra that makes both projections onto $L_\infty(\mu)$ measurable. The diagonal $D=\{(x,x)\mid x\in L_\infty(\mu)\}$ is not a measurable subset of $\Omega$: If it were then there would be a countable family of subsets of $L_\infty(\mu)$ separating the points of $L_\infty(\mu)$, so the cardinality of $L_\infty(\mu)$ would be at most continuum.

Define $f(x_0,x_1)=e^{x_0}$, $g(x_0,x_1)=e^{-x_1}$ for $(x_0,x_1)\in\Omega$. The function $fg=e^{x_0-x_1}$ is not measurable because the set $\{I\}$ is closed and its preimage is $D$.