Given two nontrivial intervals $I$ and $J$ (both open or both closed), do there always exist a monotone bijection between $I\cap \mathbb{Q}$ and $J\cap \mathbb{Q}$?
If the endpoints of $I$ and $J$ are rational numbers, then such a bijection is easy to find (just take the linear function that sends the endpoints of $I$ to those of $J$). But in general, it's not clear what to do.
On
The answer is almost yes: we need to go a little bit deeper than just closed/open, but not much deeper.
It turns out to be a bit clarifying to talk about linear orders in general, rather than specifically sets of real numbers. (Note that "order-preserving bijection" now is just "isomorphism.")
The key result is an old theorem of Cantor:
Any two countable dense linear orders without endpoints are isomorphic.
The proof is by what is now known as a "back-and-forth construction," which is fundamental in model theory.
This immediately tells us:
If $a<b$ and $c<d$ are real numbers, then $(a,b)\cap\mathbb{Q}$ and $(c,d)\cap\mathbb{Q}$ are in order-preserving bijection.
This is because clearly each interval is countable, and since they are open intervals neither has endpoints, so we can apply Cantor's theorem.
An immediate corollary of Cantor's theorem is that we can "add endpoints to both sides:" any two countable dense linear orders with left endpoints but no right endpoints are isomorphic, any two countable dense linear orders with right endpoints but no left endpoints are isomorphic, and any two countable dense linear orders with both left and right endpoints are isomorphic.
However, this doesn't immediately lift to the situation of intervals with endpoints since we now need to distinguish between rational and irrational endpoints. So for example, if $a<b$ and $c<d$ are all rational then $[a,b]\cap\mathbb{Q}$ and $[c,d]\cap\mathbb{Q}$ are in order-preserving bijection, but $[0,1]\cap\mathbb{Q}$ and $[0,\pi]\cap\mathbb{Q}$ are not in order-preserving bijection.
Ultimately - ignoring the degenerate intervals $[a,a]$ - we get four "types." Two (non-degenerate) intervals yield the same linear order when intersected with the rationals iff they are of the same "type." Specifically, a nontrivial interval $I$ is:
Type $1$ iff it contains neither a least nor greatest rational.
Type $2$ iff it contains a least rational but no greatest rational.
Type $3$ iff it contains a greatest rational but no least rational.
Type $4$ iff it contains both a least and greatest rational.
Noah Schweber's answer, using Cantor's theorem, was also the first answer that occurred to me, but here's an alternative approach that doesn't need Cantor's theorem. I'll use Noah's convenient list of four types of intervals of rationals.
Type 4, where both endpoints are in the interval (and are therefore rational), is the easiest. The linear, increasing function that sends the endpoints of one such interval to the endpoints of the other has rational coefficients and thus gives the desired bijection.
Now consider type 2, where the interval contains its left endpoint $a$ but not its right endpoint (so $a$ is rational but we don't know about $b$). We can chop up this interval $[a,b)\cap\mathbb Q$ (where I'm using the standard convention that "[" or "]" means to include the endpoint and "(" or ")" means to exclude it) into a sequence of intervals of type 4 as follows. Choose an increasing sequence $a_0,a_1,a_2,\dots$ of rational numbers with $a_0=a$ and with $\lim_{n\to\infty}a_n=b$. Then $[a,b)$ is the union of the intervals $[a_n,a_{n+1}]\cap\mathbb Q$. Now if we're given a second interval of type 2, say $[a',b')\cap\mathbb Q$, chop it up similarly into intervals of type 4, $[a_n',a_{n+1}']\cap\mathbb Q$. Then use the result already proved for type 4 to monotonically biject each $[a_n,a_{n+1}]\cap\mathbb Q$ to the corresponding $[a_n',a_{n+1}']\cap\mathbb Q$. All those bijections together constitute an increasing bijection from $[a,b)\cap\mathbb Q$ to $[a',b')\cap\mathbb Q$.
Type 3 is handled analogously, with a decreasing sequence of $a_n$'s approaching the excluded left endpoint. Finally, for type 1, pick a rational number $q$ in the open interval $(a,b)$ and break the interval into $(a,q]\cap\mathbb Q$ and $[q,b)\cap\mathbb Q$. These are of types 3 and 2 respectively, so you can handle them by the method above.