No, it is not true.

It suffices to show that the complement is not necessarily Borel. Let $B=X\setminus A$, which is a general Borel set. We have $$ \{x:H_0x\not\subseteq A\} = \{x:H_0x\cap B\ne\varnothing\} = \{x: (\exists h\in H_0)(hx\in B)\} = \{x: x\in H_0^{-1}B\} = H_0^{-1}B $$ Let $G=H_0^{-1}$ which is again a general Borel set since $G^{-1}=H_0$. Then we want to determine whether $GB$ must be Borel. But let the group be addition on $\mathbb R$: Erdös and Stone showed that the sum of Borel sets is not necessarily Borel.

No, it is not true.

It suffices to show that the complement is not necessarily Borel. Let $B=X\setminus A$, which is a general Borel set. We have $$ \{x:H_0x\not\subseteq A\} = \{x:H_0x\cap B\ne\varnothing\} = \{x: (\exists h\in H_0)(hx\in B)\} = \{x: x\in H_0^{-1}B\} = H_0^{-1}B $$ Let $G=H_0^{-1}$ which is again a general Borel set since $G^{-1}=H_0$. Then we want to determine whether $GB$ must be Borel. But let the group be addition on $\mathbb R$: Erdös and Stone showed that the sum of Borel sets is not necessarily Borel.