No. Here's an example. Let $X=\Bbb C^2$, $\hat{X}$ the blow-up of $X$ at the origin, $U=(\Bbb C^*)^2$, and $M=\mathcal{O}_U$. Then $j_{!*}M=\mathcal{O}_X$ and $i_{!*}M=\mathcal{O}_{\hat{X}}$.
Claim: $\phi_+\mathcal{O}_{\hat{X}}\ncong \mathcal{O}_X$.
Proof. Let $E=\phi^{-1}(0)$, and let $i_E\colon E\hookrightarrow \hat{X}$ and $i_0\colon \{0\} \hookrightarrow X$ be inclusion. It suffices to show that $i_0^+\mathcal{O}_X\ncong i_0^+\phi_+\mathcal{O}_{\hat{X}}$.[1] First, we have $$i_0^+\mathcal{O}_X = Li_0^*\mathcal{O}_X[-2] = \Bbb{C}[-2].$$ Next, denoting by $a_E$ the map from $E$ to a point, we have $$i_0^+\phi_+\mathcal{O}_{\hat{X}} = (a_E)_+ i_E^+\mathcal{O}_{\hat{X}} = (a_E)_+ \mathcal{O}_E[-1],$$ where the first equality is via base change. But $(a_E)_+\mathcal{O}_E[-1]$ is (up to shift) the de Rham cohomology of $E\cong \Bbb P^1$. In particular, it has cohomology in more than one degree. This finishes the proof of the claim.
[1] Given a morphism $f\colon X\to Y$ of smooth varieties, I denote by $f^+$ the (shifted) $D$-module inverse image functor. I.e. $$f^+:=Lf^*[\dim X - \dim Y].$$
