Are the real numbers the unique Dedekind-complete ordered set?
No, not at all. First of all, there are some trivial examples: you could take the empty set, or a singleton set, or a closed (or half-open) interval in $\mathbb{R}$. To eliminate such trivialities, you should additional ask for your ordered set to be nonempty and have no greatest or least element.
However, there are also nontrivial examples that cannot be ruled out so easily. Indeed, every totally ordered set has a Dedekind-completion which can be constructed just like $\mathbb{R}$ as the completion of $\mathbb{Q}$. So for instance, if you start with any dense ordered set $X$ of cardinality greater than $2^{\aleph_0}$ with no greatest or least element, its Dedekind-completion is a Dedekind-complete dense ordered set with no greatest or least element which cannot be isomorphic to $\mathbb{R}$ since it does not even have the same cardinality.
(For an explicit example of such an $X$, note that if $Y$ is any totally ordered set (e.g., an ordinal), then $Y\times\mathbb{Q}$ with the lexicographic order is a dense totally ordered set with no greatest or least element.)
Incidentally, you might wonder why you can't construct counterexamples for fields in the same way. After all, if you have any ordered field $X$, you can take its Dedekind-completion as an ordered set. You could then try to imitate the construction of the field structure on $\mathbb{R}$ from that of $\mathbb{Q}$ to get a field structure on the completion of $X$. However, it turns out that the verification of some of the field axioms on $\mathbb{R}$ uses more than just that it is the completion of an ordered field $\mathbb{Q}$--it also uses the fact that $\mathbb{Q}$ is Archimedean. So, you can similarly construct a completion of any Archimedean ordered field, but that will always just give you $\mathbb{R}$.