This holds for an arbitrary reflexive Banach space $X$ (essentially by definition of the Pettis integral). This follows from the closed graph theorem (Baire to the rescue!). Let me first treat the Dunford integral on an arbitrary Banach space (it takes values in the bidual of $X$) and then specialize to the Pettis integral later on. Reflexivity is then only used to ensure that a weakly measurable function is actually Pettis integrable.
Added: The argument I'm presenting here is quite standard, but I should point out that I learned about it in R.A. Ryan's book Introduction to tensor products on Banach spaces.
So assume that $(\Omega,\mu)$ is a ($\sigma$-finite) measure space, that $f: \Omega \to X$ is weakly $\mu$-integrable, that is: for all $\phi \in X^{\ast}$ the function $\phi \circ f$ is ($\mu$-measurable and) integrable. We then have a linear map $T_{f}: X^{\ast} \to L^{1}(\Omega,\mu)$, which I claim to be bounded. So suppose that $\phi_{n} \to \phi$ in $X^{\ast}$ and $T_f(\phi_{n}) \to g$ in $L^{1}$. Passing to a subsequence, we may assume that $T_f(\phi_{n}) \to g$ a.e. On the other hand, $\phi_{n} \circ f \to \phi \circ f$ pointwise everywhere, so $\phi \circ f = g$ a.e., in other words $T_f \phi = g$ in $L^1$, so $T_{f}$ is indeed bounded by the closed graph theorem.
Now the adjoint operator $T^{\ast}: L^{\infty} \to X^{\ast\ast}$ satisfies for $h \in L^{\infty}$ $ \langle \phi, T_{f}^{\ast}h \rangle = \int_{\Omega} h \cdot T_f\phi = \int_{\Omega} h \cdot \phi \circ f,$ where I used $\sigma$-finiteness in order to ensure that the duality $(L^{1})^{\ast} = L^{\infty}$ works nicely.
If $E \subset \Omega$ is measurable, then $T_{f}^{\ast} \chi_{E} =: \int_{E} f\,d\mu \in X^{\ast\ast}$ is called the Dunford integral of $f$ over $E$ and it is a tautology that every weakly $\mu$-integrable function $f: \Omega \to X$ has a Dunford integral.
In general, the function $f$ is called Pettis-integrable if $T_{f}^{\ast}\chi_{E} \in X \subset X^{\ast\ast}$ for every measurable $E \subset \Omega$. If $f$ is Pettis-integrable then $T_{f}^{\ast}$ actually defines an operator $S_{f}: L^{\infty} \to X$, as the characteristic functions span a norm-dense subspace of $L^{\infty}$ and $X$ is closed in $X^{\ast\ast}$. Now what you call the Pettis integral of $f$, is simply $S_{f} \chi_{\Omega} = T_{f}^{\ast} \chi_{\Omega} = \int_{\Omega} f\,d\mu \in X$.
Finally, if $X$ is reflexive, then every weakly $\mu$-integrable $f: \Omega \to X$ is Pettis integrable (by definition of reflexivity), and of course, putting $y = \int_\Omega f\,d\mu = T_{f}^{\ast}\chi_{\Omega} \in X$ we get the identity and existence you asked about for free: for $\phi \in X^{\ast}$ we have $\langle \phi, y \rangle_{X^\ast, X} = \langle \phi, T_{f}^{\ast}\chi_{\Omega}\rangle_{X,X^{\ast\ast}} = \int_{\Omega} \chi_{\Omega} \cdot \phi \circ f\,d\mu = \int_{\Omega} \langle \phi, f(x)\rangle_{X^{\ast},X}\,d\mu$ as we wanted.