I don't think $\pi_\omega(A)''$ has that form. For example, take $A = M_2$ and let $\omega$ be the normalized trace. Then $\omega = \frac{1}{2}(\psi_1 + \psi_2)$ where $\psi_i(x) = \langle xe_i, e_i\rangle$, for $x \in M_2$ and $\{e_1,e_2\}$ the standard basis of $\mathbb{C}^2$. That is, $\omega$ is the integral $\int \psi\, d\mu(\psi)$ where $\mu = \frac{1}{2}(\delta_{\psi_1} + \delta_{\psi_2})$. Then $\pi_\omega(M_2)'' \cong \pi_{\psi_1}(M_2)'' \cong \pi_{\psi_2}(M_2)'' \cong M_2$, so $\pi_\omega(M_2)'' \not\cong \pi_{\psi_1}(M_2)'' \oplus \pi_{\psi_2}(M_2)''$.

We do have an isomorphism between $H_\omega$ and $H_{\psi_1} \oplus H_{\psi_2}$ which takes $\pi_\omega(x)$ to $\pi_{\psi_1}(x) \oplus \pi_{\psi_2}(x)$, but of course that doesn't make $\pi_\omega(A) \cong \pi_{\psi_1}(A) \oplus \pi_{\psi_2}(A)$.