You're right; this is an error in what you've heard. The embedding $i:P\to \hat P$ of a poset $P$ into its complete poset of downsets $\hat P$ is the free completion of the poset, in the sense that every poset morphism
$P\to Q$, where $Q$ is complete, has a unique factorization through $i$ preserving all joins. However, $[-\infty,\infty]$ is not quite the free completion of $\mathbb{Q}$. Indeed, there is no factorization of $i:\mathbb Q\to \widehat{\mathbb Q}$ through the inclusion $\mathbb{Q}\to [-\infty,\infty]$ preserving joins, since the join of the set $\{1-1/n:n\in\mathbb N\}$ in $[-\infty,\infty]$ is $1$, while in $\widehat{\mathbb Q}$ it is $\{x:x<1\}$.
The property that $[-\infty,\infty]$ does have, and more generally that characterizes Dedekind-MacNeille completions $j:P\to \bar P$ of posets, is that $[-\infty,\infty]$ is the universal complete poset admitting a map from $\mathbb{Q}$ which preserves joins which already exist. This is in contrast to $i:\mathbb Q\to \widehat{\mathbb Q}$, which destroys the join mentioned at the end of the last paragraph.
Indeed free cocompletions destroy whatever colimits in a given enriched setting are not absolute-joints of finite sets in posets, splitting of idempotents in ordinary categories, biproducts in categories enriched over abelian groups, etc. So if $Q$ is some complete poset, then maps $\mathbb Q\to Q$ which preserve joins factor uniquely through $[-\infty,\infty]$. We could say that $[-\infty,\infty]$ is the free conservative cocompletion of $\mathbb{Q}$-the cocompletion that doesn't mess up anything we've already handled.
Arguably the Dedekind completion is a bit more abstruse, categorically, than the Cauchy completion. If one views $\mathbb{Q}$ as a category enriched over $[0,\infty]$, that is, as a generalized metric space, then the Cauchy completion is just the closure of $\mathbb{Q}$ in its category of enriched presheaves under absolute colimits-the absolute colimits in metric spaces being essentially limits of Cauchy sequences. For this reason, the closures of other kinds of enriched categories under absolute colimits are sometimes also called Cauchy completions.
We can string together definitions adjunctions, and the occasional use of the Yoneda lemma to get (for any object $C$ in the category $\mathbb{C}$):
\begin{align*}
y(B)^{y(A)}(C)
&\cong \operatorname{Hom}(y(C), y(B)^{y(A)}) \\
&\cong \operatorname{Hom}(y(C) \times y(A), y(B)) \\
&\cong \operatorname{Hom}(y(C \times A), y(B)) \\
&\cong y(B)(C \times A) \\
&= \operatorname{Hom}(C \times A, B) \\
&\cong \operatorname{Hom}(C, B^A) \\
&= y(B^A)(C) \,.
\end{align*}
Best Answer
You can recover a category $\mathcal{C}$ from its category of presheaves, up to Cauchy completion. In particular that means that if $\mathcal{C}$ is Cauchy complete, then you can recover it (up to equivalence) from its category of presheaves. A lot of information about this can be found on the nLab page about Cauchy complete categories.
Let me just state the major relevant points here. The numbering in this answer refers to the (current) numbering on the nLab page.
Note that if a category has equalizers, then it is Cauchy complete (take $i$ to be the equalizer of $e$ and $Id_C$, and $r$ the universal arrow corresponding to $e$).
Then Proposition 2.3 states that if $\mathcal{C}$ is Cauchy complete, then it can be recovered from $\mathbf{Set}^{\mathcal{C}^\text{op}}$ as the full subcategory of "tiny objects" or "small projective objects".
There are equivalent characterisations possible. For example, we could also recover $\mathcal{C}$ as the indecomposable projectives.
If $\mathcal{C}$ is not Cauchy complete, then taking the tiny objects will give us the Cauchy completion $\bar{\mathcal{C}}$. We cannot hope to do better, because there is an equivalence of categories $$ \mathbf{Set}^{\mathcal{C}^\text{op}} \simeq \mathbf{Set}^{\bar{\mathcal{C}}^\text{op}}. $$