Okay after thinking about this again I think I know why this is true. Since $\Sigma$ is Fiberwise starshaped then the Liouville form $\lambda$ of the cotangent bundle will be a contact form on $\Sigma$, and we will have that $d\lambda=\omega$ restricted to $T\Sigma$ will have rank $2n-2$ and it's kernel will be $1$-dimensional. Now by definition of the Hamiltonian vector fields we will have that $0=dH(.)|_{T\Sigma}=\omega(X_H|_{\Sigma},.)$ and the same happends with $X_G|_{\Sigma}$ and so since the kernel in $T\Sigma$ is one-dimensional we have that there exits a function $\sigma(x)$ such that $X_H(x)=\sigma(x)X_G(x)$ for $x\in \Sigma$. And so if $\phi_G^t$ is the Hamiltonian flow of $G$ we have that $\phi_G^{f(t,x)}$ , where $f(t,x)=\int_{0}^{t}\sigma(x)dx$ is the hamiltonian flow of $H$ since $\dot \phi_G^{f(t,x)}=\dot f(t,x)X_G(x)=\sigma(x)X_G(x)=X_H(x)$.

Okay after thinking about this again I think I know why this is true. Since $\Sigma$ is Fiberwise starshaped then the Liouville form $\lambda$ of the cotangent bundle will be a contact form on $\Sigma$, and we will have that $d\lambda=\omega$ restricted to $T\Sigma$ will have rank $2n-2$ and it's kernel will be $1$-dimensional. Now by definition of the Hamiltonian vector fields we will have that $0=dH(.)|_{T\Sigma}=\omega(X_H|_{\Sigma},.)$ and the same happends with $X_G|_{\Sigma}$ and so since the kernel in $T\Sigma$ is one-dimensional we have that there exits a smooth function $\sigma(x)$ such that $X_H(x)=\sigma(x)X_G(x)$ for $x\in \Sigma$. And so if $\phi_G^t$ is the Hamiltonian flow of $G$ we have that $\phi_G^{f(t,x)}$ , where $f(t,x)=\int_{0}^{t}\sigma(x)dx$ is the hamiltonian flow of $H$ since $\dot \phi_G^{f(t,x)}=\dot f(t,x)X_G(x)=\sigma(x)X_G(x)=X_H(x)$.