I think you do not need any extra set theoretic assumptions beyond ZFC. (In particular you do not need the continuum hypothesis, and you do not need Martin's axiom.)
Repeat the usual construction of a Bernstein set: Let $\{F_\alpha: \alpha < \mathfrak c = 2^{\aleph_0}\}$ be a list of all closed uncountable subsets of the real line. Call a set $S$ free if it contains no three different points $x,y,z$ with $y-x=z-y$ (that is, no points with $y=\frac{x+z}2$). Let $E(S)=\{z\in\mathbb R\setminus S: \exists x,y\in S, z=\frac{x+y}2,\mathrm{\ or\ } z=y+(y-x)\}$, that is $E(S)$ are the excluded points for $S$. (Note that $z=y+(y-x)$ could be to the left or to the right of $y$, depending on whether $y-x$ is negative or positive.) For each $\alpha$ pick different $x_\alpha,y_\alpha\in F_\alpha$ such that if $P_\alpha=\{x_\beta:\beta<\alpha\}$ and $Q_\alpha=\{y_\beta:\beta<\alpha\}$ then we have :
$x_\alpha\not\in E(P_\alpha)\cup P_\alpha\cup Q_\alpha$, and $y_\alpha\not\in P_\alpha\cup Q_\alpha$.
Let $P=\{x_\alpha:\alpha<\mathfrak c\}$ and $Q=\mathbb R\setminus P$. Then each of $P$ and $Q$ is a Bernstein set of inner measure zero and full outer measure. Moreover, $P$ is free. We have that $X\cap P=X\setminus Q$ and since $Q$ is of inner measure zero, the outer measure $m^*(X)=m^*(X\setminus Q)$. That is $m^*(X\cap P)=m^*(X)$, and $X\cap P$ is free since $P$ is.