Not necessarily, unless $H$ is a normal subgroup.
For example, consider the group $G=S_3=\langle g,h|g^3=h^2=ghgh=e\rangle=\{e,g,g^2,h,gh,g^2h\}$, and let $H=\langle h\rangle=\{e,h\}$. It's a pretty easy multiplication table to construct (if you need to). List out the left cosets and right cosets of $H$, and you'll find several examples of $x,y\in G$ such that $xH\neq Hy$ and $xH\cap Hy\neq \emptyset$.
Note however that there are the same number of left cosets as right cosets. This holds in general for any group $G$ and any subgroup $H$. Try showing that $xH\mapsto Hx^{-1}$ is a (well-defined) bijection from the left cosets of $H$ in $G$ to the right cosets of $H$ in $G$. That is, show that $xH=yH$ if and only if $Hx^{-1}=Hy^{-1}$ (well-defined and injective), and observe that surjectivity is simple since $G$ is a group. (Hint for the first part: Don't forget that $H$ is a group!)