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Warning: The following answer has a gap, and I do not know if it could be fixed. I leave it for consideration, now that I already wrote it. (It does work, in ZFC, but under the additional assumption that $X$ is measurable.)

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 (or "forbidden") 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 $y_\alpha\not\in P_\alpha\cup Q_\alpha$.
(We could do this since $|F_\alpha|=\mathfrak c$, but $|E(P_\alpha)|\le\aleph_0\cdot |P_\alpha|\le\aleph_0\cdot|\alpha|<\mathfrak c$.)

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, by construction $P$ is free. We have that $X\cap P=X\setminus Q$ and since $Q$ is of inner measure zero, the outer measure $m^*(X)=m^*(X\setminus Q)$ (this is the gap in my proof, this is not necessarily the case, unless we assume that $X$ is measurable). That is $m^*(X\cap P)=m^*(X)$, and $X\cap P$ is free since $P$ is.

Well, the gap is, I only know that $X$ has positive outer measure, but this does not rule out the possibility that $X$ is of inner measure zero, so it might happen that $X\subset Q$ and in that case $X\cap P$ would be empty.

I got mislead by the following answer (even if the answer is fine, but I didn't read it right, it only applies when we start with a measurable set, and the set $X$ above is not necessarily measurable):
A set $E$ with positive Lebesgue measure can be decomposed as a union $E=A\cup B$ where each of $A$ and $B$ have zero inner measure, and therefore each of $A$ and $B$ are nonmeasurable with $m^*(A)=m^*(B)=m(E)$.
Non Lebesgue measurable subsets with "large" outer measure

Warning: The following answer has a gap, and I do not know if it could be fixed. I leave it for consideration, now that I already wrote it. (It does work, in ZFC, but under the additional assumption that $X$ is measurable.)

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 (or "forbidden") 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 $y_\alpha\not\in P_\alpha\cup Q_\alpha$.
(We could do this since $|F_\alpha|=\mathfrak c$, but $|E(P_\alpha)|\le\aleph_0\cdot |P_\alpha|\le\aleph_0\cdot|\alpha|<\mathfrak c$.)

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, by construction $P$ is free. We have that $X\cap P=X\setminus Q$ and since $Q$ is of inner measure zero, the outer measure $m^*(X)=m^*(X\setminus Q)$ (this is the gap in my proof, this is not necessarily the case, unless we assume that $X$ is measurable). That is $m^*(X\cap P)=m^*(X)$, and $X\cap P$ is free since $P$ is.

Well, the gap is, I only know that $X$ has positive outer measure, but this does not rule out the possibility that $X$ is of inner measure zero, so it might happen that $X\subset Q$ and in that case $X\cap P$ would be empty.

I got mislead by the following answer (even if the answer is fine, but I didn't read it right, it only applies when we start with a measurable set, and the set $X$ above is not necessarily measurable):
A set $E$ with positive Lebesgue measure can be decomposed as a union $E=A\cup B$ where each of $A$ and $B$ have zero inner measure, and therefore each of $A$ and $B$ are nonmeasurable with $m^*(A)=m^*(B)=m(E)$.
Non Lebesgue measurable subsets with "large" outer measure

Warning: The following answer has a gap, and I do not know if it could be fixed. I leave it for consideration, now that I already wrote it. (It does work under the additional assumption that $X$ is measurable.)

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 $y_\alpha\not\in P_\alpha\cup Q_\alpha$.
(We could do this since $|F_\alpha|=\mathfrak c$, but $|E(P_\alpha)|\le\aleph_0\cdot |P_\alpha|\le\aleph_0\cdot|\alpha|<\mathfrak c$.)

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, by construction $P$ 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.

Well, the gap is, I only know that $X$ has positive outer measure, but this does not rule out the possibility that $X$ is if inner measure zero, so it might happen that $X\subset Q$ and in that case $X\cap P$ would be empty.

I got mislead of the following answer (even if the answer is fine, but I didn't read it right, it only applies when we start with a measurable set, and the set $X$ above is not necessarily measurable):
A set $E$ with positive Lebesgue measure can be decomposed as a union $E=A\cup B$ where each of $A$ and $B$ have zero inner measure, and therefore each of $A$ and $B$ are nonmeasurable with $m^*(A)=m^*(B)=m(E)$.

 Non Lebesgue measurable subsets with "large" outer measure

Warning: The following answer has a gap, and I do not know if it could be fixed. I leave it for consideration, now that I already wrote it. (It does work, in ZFC, but under the additional assumption that $X$ is measurable.)

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 (or "forbidden") 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 $y_\alpha\not\in P_\alpha\cup Q_\alpha$.
(We could do this since $|F_\alpha|=\mathfrak c$, but $|E(P_\alpha)|\le\aleph_0\cdot |P_\alpha|\le\aleph_0\cdot|\alpha|<\mathfrak c$.)

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, by construction $P$ is free. We have that $X\cap P=X\setminus Q$ and since $Q$ is of inner measure zero, the outer measure $m^*(X)=m^*(X\setminus Q)$ (this is the gap in my proof, this is not necessarily the case, unless we assume that $X$ is measurable). That is $m^*(X\cap P)=m^*(X)$, and $X\cap P$ is free since $P$ is.

Well, the gap is, I only know that $X$ has positive outer measure, but this does not rule out the possibility that $X$ is of inner measure zero, so it might happen that $X\subset Q$ and in that case $X\cap P$ would be empty.

I got mislead by the following answer (even if the answer is fine, but I didn't read it right, it only applies when we start with a measurable set, and the set $X$ above is not necessarily measurable):
A set $E$ with positive Lebesgue measure can be decomposed as a union $E=A\cup B$ where each of $A$ and $B$ have zero inner measure, and therefore each of $A$ and $B$ are nonmeasurable with $m^*(A)=m^*(B)=m(E)$. 
Non Lebesgue measurable subsets with "large" outer measure

Warning: The following answer has a gap, and I do not know if it could be fixed. I leave it for consideration, now that I already wrote it.

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 $y_\alpha\not\in P_\alpha\cup Q_\alpha$.
(We could do this since $|F_\alpha|=\mathfrak c$, but $|E(P_\alpha)|\le\aleph_0\cdot |P_\alpha|\le\aleph_0\cdot|\alpha|<\mathfrak c$.)

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, by construction $P$ 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.

Well, the gap is, I only know that $X$ has positive outer measure, but this does not rule out the possibility that $X$ is if inner measure zero, so it might happen that $X\subset Q$ and in that case $X\cap P$ would be empty.

I got mislead of the following answer (even if the answer is fine, but I didn't read it right, it only applies when we start with a measurable set, and the set $X$ above is not necessarily measurable):
A set $E$ with positive Lebesgue measure can be decomposed as a union $E=A\cup B$ where each of $A$ and $B$ have zero inner measure, and therefore each of $A$ and $B$ are nonmeasurable with $m^*(A)=m^*(B)=m(E)$.

Non Lebesgue measurable subsets with "large" outer measure

Warning: The following answer has a gap, and I do not know if it could be fixed. I leave it for consideration, now that I already wrote it. (It does work under the additional assumption that $X$ is measurable.)

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 $y_\alpha\not\in P_\alpha\cup Q_\alpha$.
(We could do this since $|F_\alpha|=\mathfrak c$, but $|E(P_\alpha)|\le\aleph_0\cdot |P_\alpha|\le\aleph_0\cdot|\alpha|<\mathfrak c$.)

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, by construction $P$ 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.

Well, the gap is, I only know that $X$ has positive outer measure, but this does not rule out the possibility that $X$ is if inner measure zero, so it might happen that $X\subset Q$ and in that case $X\cap P$ would be empty.

I got mislead of the following answer (even if the answer is fine, but I didn't read it right, it only applies when we start with a measurable set, and the set $X$ above is not necessarily measurable):
A set $E$ with positive Lebesgue measure can be decomposed as a union $E=A\cup B$ where each of $A$ and $B$ have zero inner measure, and therefore each of $A$ and $B$ are nonmeasurable with $m^*(A)=m^*(B)=m(E)$.

Non Lebesgue measurable subsets with "large" outer measure