Normally, the graph will not be distance-regular.

Pick your favorite (but not too small) set $S\subset\mathbb Z_2^n$. Unless you are very unlucky, there will be two elements $s_1,s_2\in S$ which are both representable as a sum of two (other) elements of $S$, but the number of representations differ: say, $$ s_1=t_1'+t_1'',\ s_2=t_2+t_2'',\quad t_1',t_1'',t_2',t_2''\in S, $$ while $r_{2S}(s_1)\ne r_{2S}(s_2)$. Now, $t_1'$ and $t_1''$ are distance $1$ apart in the graph under consideration, and so are $t_2'$ and $t_2''$. However, the number of common neighbors of $t_1'$ and $t_1''$ is distinct from the number of common neighbors of $t_2'$ and $t_2''$ (the former is $r_{2S}(s_1)$, the latter is $r_{2S}(s_2)$).

Normally, the graph will not be distance-regular.

Pick your favorite (but not too small) set $S\subset\mathbb Z_2^n$. Unless you are very unlucky, there will be two elements $s_1,s_2\in S$ which are both representable as a sum of two (other) elements of $S$, but the number of representations differ: say, $$ s_1=t_1'+t_1'',\ s_2=t_2'+t_2'',\quad t_1',t_1'',t_2',t_2''\in S, $$ while $r_{2S}(s_1)\ne r_{2S}(s_2)$. Now, $t_1'$ and $t_1''$ are distance $1$ apart in the graph under consideration, and so are $t_2'$ and $t_2''$. However, the number of common neighbors of $t_1'$ and $t_1''$ is distinct from the number of common neighbors of $t_2'$ and $t_2''$ (the former is $r_{2S}(s_1)$, the latter is $r_{2S}(s_2)$).