The answer to question 2 (and therefore question 1 as well) is no. This follows from the Lebesgue Density Theorem: if $\mathbb Q + D$ is measurable, then for every open interval $(a,b)$, $\frac{1}{b-a}\mu(\mathbb Q + D)$ must be the same, and then the Lebesgue Density Theorem implies that it must always be either $0$ or $1$.

The answer to question 2 (and therefore question 1 as well) is no. This follows from the Lebesgue Density Theorem, which says, roughly, that a measurable set that is neither null nor co-null cannot be too evenly spread out, but it must be "lumpy" in places, like a fat Cantor set. If $\mathbb Q + D$ is measurable, then for every open interval $(a,b)$, $$\frac{1}{b-a}\mu((a,b) \cap (\mathbb Q + D))$$ must be the same, and then the Lebesgue Density Theorem implies that it must always be either $0$ or $1$.