Yes. We proceed using the axiom of choice. Choose an basis of $\mathbb{R}$ with cardinality $|\mathbb{R}/\mathbb{Q}|$ and a bijection between this basis and the elements of $\mathbb{R}/\mathbb{Q}$. Pick an element from each coset of $\mathbb{Q}\subset \mathbb{R}$$\mathbb{Q}$ contained in the corresponding basis set (we may do this as each coset is dense); this will be our set $D$. It is dense as each basis set has non-empty intersection with $D$. Now any sequence of rationals tending to zero works as $c(n)$, as if $D+c(n)$ had non-empty intersection with $D$, then two elements of $D$ would differ by a rational.
