$M_{\phi_1}(\Sigma_g)$ and $M_{\phi_2}(\Sigma_g)$ are the same Seifert fibered space, denoted as $M$. Let $\zeta_1,\zeta_2$ be the cohomology classes of $M$ representing the surface bundle structures (i.e. dual to $\Sigma_g$). It suffices to find a self-homeomorphism of $M$ that pulls $\zeta_1$ back to $\zeta_2$.
The base orbifold of $M$, denoted as $B$, is orientable and has orbifold Euler characteristic $\chi(B)<0$, and the rational Euler number $e(M)=0$. Let $h$ denote the homology class of $M$ representing a regular fiber. Then, $\chi(\Sigma_g)=|\zeta_i(h)|\chi(B)$, so $\zeta_1(h)=\pm \zeta_2(h)$. Up to a self-homeomorphism of $M$ that reverses the direction of the fiber, we may assume $\zeta_1(h)=\zeta_2(h)=n.$
Let $\check{B}$ denote the surface obtained from $B$ by removing the open neighborhoods of the cone points. We view $M$ as a Dehn surgery of $\check{B}\times S^1$ so that we fix an embedding $\check{B}\to M$.
A cohomology class $\zeta\in H^1(M,\mathbb Z)$ is completely determined by $\zeta|_{\check{B}}\in H^1(\check{B},\mathbb{Z})$ and $\zeta(h)$. Since $\zeta(h)$ determines $\zeta|_{\partial\check{B}}$, we have $\zeta_1|_{\partial\check{B}}=\zeta_2|_{\partial\check{B}}$. Let $\zeta_i|_{\check{B}}^{\mathrm{mod}\,n}$ denote the image of $\zeta_i|_{\check{B}}$ in $H^1(\check{B},\mathbb{Z}/n\mathbb{Z})$. Since $\zeta_i$ is a primitive class, i.e. surjective onto $\mathbb{Z}$, $\zeta_i|_{\check{B}}^{\mathrm{mod}\,n}$ must also be a primitive class, i.e. surjective onto $\mathbb{Z}/n\mathbb{Z}$. Then, there exists (?) a self-homeomorphism $f$ of $\check{B}$ fixing pointwise $\partial \check{B}$ that sends $\zeta_1|_{\check{B}}^{\mathrm{mod}\,n}$ to $\zeta_2|_{\check{B}}^{\mathrm{mod}\,n}$. The self-homeomorphism $(f,id_{S^1})$ on $\check{B}\times S^1$ extends to a self-homeomorphism $F$ on $M$ such that $F^\ast\zeta_1(h)=n$, $F^\ast\zeta_1|_{\partial \check{B}}=\zeta_1|_{\partial \check{B}}=\zeta_2|_{\partial \check{B}}$, and $F^\ast\zeta_1|_{\check{B}}\equiv\zeta_2|_{\check{B}}\pmod{n}$. Then, one can composite some vertical Dehn twists in the interior of $\check{B}\times S^1$, denoted as $T$, such that $T^\ast F^\ast \zeta_1|_{\check{B}}=\zeta_2|_{\check{B}}$, and we are done.
Let me explain how the vertical Dehn twists work. Let $\alpha_1,\beta_1,\cdots,\alpha_{g},\beta_g$ be the standard simple closed curves on $\check{B}$ which form a free basis of the first homology of the closed surface obtained from capping off $\partial \check{B}$ with disks, and whose intersection form is $\mathrm{diag}\left\{\left ( \begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix}\right), \cdots, \left ( \begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix}\right)\right\}.$ Let $t$ be a vertical Dehn twist supported in a neighborhood of $\alpha_1\times S^1$. Then $t_\ast [\alpha_1]=[\alpha_1]$, $t_\ast[\beta_1]=[\beta_1]+[h]$, and $t_\ast$ fixes $[\alpha_2],[\beta_2],\cdots,[\alpha_g],[\beta_g],[h]$. So $t^\ast \zeta(\beta_1)=\zeta(\beta_1)+n$, where $n=\zeta(h)$, and all other values are fixed.
