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 me give two proofs showing that $L$ has a copy of $\mathbb{Q}$.

First, we'll build it 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 $\mathbb{Q}$.

Alternatively, we may argue via the Cantor-Bendixson analysis. Consider $L$ in the order topology, and form the Cantor-Bendixson derivatives, by casting out isolated points at each step and taking intersections at limits. $$L_0=L\\ L_{\alpha+1}=L_\alpha'\\ L_\lambda=\bigcap_{\alpha<\lambda}L_\alpha.$$ In countably many steps, the process terminates in either the empty set or in a suborder of $L$ with no isolated points. Since isolated points do not add any new cuts to the Dedekind completion, it follows that the process will be throwing out exactly the same points of $L$ when carried out with the completion as it does with $L$ itself (although the Cantor-Bendixson process applied to the completion of $L$ may cast out its own additional new points, as with $\mathbb{Z}+\mathbb{Z}$). Since we throw out only countably many points altogether and the completion of $L$ is uncountable, it follows that with $L$ we must not have terminated in the empty set. Thus, $L$ has a suborder with no isoated points, and therefore 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.

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 $L$.