Countable shifts of closed positive sets

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 $1$.

Another reason for "no":

Replace the numbers in the unit interval by their binary representation; Lebesgue measure now becomes the product measure on $2^\omega$. Writing $Q'$ for the set of dyadic rationals, the set $C+Q'$ is now a tail set, so it must have measure $0$ or $1$, by Kolmogorov's 0-1-law. If $C$ was positive, $C+Q'$ has measure $1$.

Easier (well, I mean: requiring less deep knowledge) reason than in the answers by Goldstern and Will Brian.

Your set $C$ may be approximated by a disjoint union of small dyadic intervals (simply disjoint intervals also work) with accuracy 1 percent. Thus by pigeonhole principle one of them consists of points of $C$ by at least 99 percents. Its disjoint (or almost disjoint) rational shifts, which form a partition of the whole segment $[0,1]$, show that $[0,1]$ consists of points of $C+\mathbb{Q}$ by at least 99 percents.