As I have written above the affimative answer itself was known to many people including T. Bartoszyński. The following proof is due to T. Weiss (my advisor).

Proof. We follow closely the proof and notation of Lemma 3.2.42 from [1]. Let $A$ be a Borel measure zero set in $M[r]$, where $r$ is a random real over $M$. There exists $\dot{A}\subseteq 2^{\omega}\times 2^{\omega}$ measure zero set coded in $M$, such that $\dot{A}_{r}=A$ (notation: $\dot{A}_{r}=\{y\colon \left<r,y\right>\in \dot{A}\}$).

Then $$\dot{A}\subseteq\bigcap_{m\in\omega}\bigcup_{n\geq m} [s_{n}]\times[t_{n}]$$ where $s_n, t_n\in 2^{<\omega}$, $\sum_{n=0}^{\infty}\frac{1}{2^{2|s_{n}|}}<\infty$ and we can assume that $|t_{n}|=|s_{n}|$ for any $n\in\omega$.

Let $z\in 2^{\omega}\cap M$ and $f\in\omega^\omega$ be increasing. Then $$\mu(\{x\colon\left<x,x_f+z\right>\in [s]\times[t]\})\leq \frac{2^{f^{-1}(|s|)}}{2^{|s|+|t|}}$$ (where $x_f\in 2^{\omega}$ such that $x_{f}(n)=x(f(n))$). By induction on length $|s_{n}|$ we define an increasing function $f_{A}\in\omega^{\omega}$ such that $$\sum_{n=0}^{\infty}\frac{2^{f^{-1}(|s_{n}|)}}{2^{2|s_{n}|}}<\infty.$$ It is easy to see that such function exists as for any $\varepsilon>0$ we can find $N_{\varepsilon}\in\omega$ such that $\sum_{n\geq N_{\varepsilon}}\frac{1}{2^{2|s_{n}|}}<\varepsilon$.

Notice also that $\left< x,x_{f}+z\right>\in [s]\times[t] $ if and only if $\left<x,x_{f}\right>\in[s]\times [t+z]$.

The set $$H_{z}=\{x\in\left<x,x_{f_A}+z\right>\in\bigcap_{m\in\omega}\bigcup_{n\geq m} [s_{n}]\times[t_{n}]\}$$ has measure zero and is coded in $M$ for every $z\in 2^{\omega}\cap M$. Thus $r\notin H_{z}$ and $\left<r,r_{f_{A}}+z\right>\notin \dot{A}$. This implies that $r_{f_{A}}\notin A+z$ for every $z\in 2^{\omega}\cap M$, so $(2^{\omega}\cap M)+A\neq 2^{\omega}$ and so $2^{\omega}\cap M$ is strongly meager.

$\square$

References

[1] T. Bartoszyński, H. Judah, Set thoery: on the structure of the real line, A K Peters, 1995

As I have written above the affimative answer itself was known to many people including T. Bartoszyński. The following proof is due to T. Weiss (my advisor).

Proof. We follow closely the proof and notation of Lemma 3.2.42 from [1]. Let $A$ be a Borel measure zero set in $M[r]$, where $r$ is a random real over $M$. There exists $\dot{A}\subseteq 2^{\omega}\times 2^{\omega}$ measure zero set coded in $M$, such that $\dot{A}_{r}=A$ (notation: $\dot{A}_{r}=\{y\colon \left<r,y\right>\in \dot{A}\}$).

Then $$\dot{A}\subseteq\bigcap_{m\in\omega}\bigcup_{n\geq m} [s_{n}]\times[t_{n}]$$ where $s_n, t_n\in 2^{<\omega}$, $\sum_{n=0}^{\infty}\frac{1}{2^{2|s_{n}|}}<\infty$ and we can assume that $|t_{n}|=|s_{n}|$ for any $n\in\omega$.

Let $z\in 2^{\omega}\cap M$ and $f\in\omega^\omega$ be increasing. Then $$\mu(\{x\colon\left<x,x_f+z\right>\in [s]\times[t]\})\leq \frac{2^{f^{-1}(|s|)}}{2^{|s|+|t|}}$$ (where $x_f\in 2^{\omega}$ such that $x_{f}(n)=x(f(n))$). By induction on length $|s_{n}|$ we define an increasing function $f_{A}\in\omega^{\omega}$ such that $$\sum_{n=0}^{\infty}\frac{2^{f_{A}^{-1}(|s_{n}|)}}{2^{2|s_{n}|}}<\infty.$$ It is easy to see that such function exists as for any $\varepsilon>0$ we can find $N_{\varepsilon}\in\omega$ such that $\sum_{n\geq N_{\varepsilon}}\frac{1}{2^{2|s_{n}|}}<\varepsilon$.

Notice also that $\left< x,x_{f}+z\right>\in [s]\times[t] $ if and only if $\left<x,x_{f}\right>\in[s]\times [t+z]$.

The set $$H_{z}=\{x\in\left<x,x_{f_A}+z\right>\in\bigcap_{m\in\omega}\bigcup_{n\geq m} [s_{n}]\times[t_{n}]\}$$ has measure zero and is coded in $M$ for every $z\in 2^{\omega}\cap M$. Thus $r\notin H_{z}$ and $\left<r,r_{f_{A}}+z\right>\notin \dot{A}$. This implies that $r_{f_{A}}\notin A+z$ for every $z\in 2^{\omega}\cap M$, so $(2^{\omega}\cap M)+A\neq 2^{\omega}$ and so $2^{\omega}\cap M$ is strongly meager.

$\square$

References

[1] T. Bartoszyński, H. Judah, Set thoery: on the structure of the real line, A K Peters, 1995