For countable linear orders, there is a nice characterization, which I see is conjectured in the comments. 

**Theorem.** The following are equivalent for a countable linear
order $\langle L,\lt\rangle$.

 1. The Dedekind completion of $L$ has size continuum.

 2. The Dedekind completion of $L$ is uncountable.

 3. $L$ contains a copy of $\mathbb{Q}$.

Proof. Clearly $3\to 1\to 2$, so it remains to see that $2\to 3$.
Suppose $L$ is countable, but has an uncountable Dedekind
completion. Let's build a copy of $\mathbb{Q}$ in $L$ directly.
Since the completion of $L$ has no
increasing or decreasing $\omega_1$ sequence, there must be two
nodes $a_0<a_1$ in $L$, such that the interval between them has an
uncountable Dedekind completion. Continuing, we may find
$a_{\frac12}$ between them, such that both intervals
$[a_0,a_{\frac12}]$ and $[a_{\frac12},a_1]$ have uncountable
Dedekind completion. Thus, by induction, we may construct a
countable dense suborder of $L$, and so $\langle L,\lt\rangle$
contains a copy of $\mathbb{Q}$. QED

In the uncountable case, things are little more complicated. For example, there are complications caused by GCH-type issues. If 
$2^\omega=2^{\omega_1}$, then already a countable suborder of a linear order $L$ of size
$\omega_1$ is sufficient to make the Dedekind completion have size $2^{\omega_1}$, since if $\mathbb{Q}$ is there the completion will have size continuum. In
general, for orders of size $\delta$, one should look at the
smallest $\kappa\leq\delta$ for which $2^\kappa>\delta$. There are some basis theorems for uncountable orders that will probably be useful.

**Edit.** I removed the previous second, alternative, proof of the theorem, which had appealed to the Cantor-Bendixson derivatives, because it doesn't actually work the way that I had claimed. For example, if $L$ is $\mathbb{Q}$ copies of $\mathbb{Z}$, then every point in $L$ is isolated, cast out on the first step, yet the Dedekind completion is still uncountable and there is still a copy of $\mathbb{Q}$ in $L$.