As you observe, every countable ordinal embeds topologically (indeed, order-theoretically) into the rationals. Distinct ordinals can be homeomorphic (e.g. $\omega+1\cong\omega+2$), but distinct ordinals of the form $\omega^\alpha$ for some $\alpha$ (= "right-indecomposable") aren't, so this gives us $\aleph_1$ many homeomorphism types. But contra your edit, this leaves open the issue of whether there are as many as possible (= $\mathfrak{c}$-many) in case CH fails.
To get continuum-many, we can use the following trick. First fix a closed $A\subset\mathbb{Q}$ of ordertype $\omega^\omega$; note that $A$ has limit points of every finite order (limit, limit-of-limits, limit-of-limits-of-limits, etc.), and let $\alpha_i$ be an element of $A$ which is an $i$-fold limit but not an $(i+1)$-fold limit for each $i$. For $p\in A$, let $\epsilon_p$ be half the distance between $p$ and the next element of $A$ (which is positive since $A$ is well-ordered).
Given $X\subseteq\mathbb{N}$, let $$A_X=A\cup\bigcup_{i\in X}([\alpha_i,\alpha_i+\epsilon_{\alpha_i}]\cap\mathbb{Q}).$$ Basically, we attach a "$\mathbb{Q}$-like label" to the $\alpha_i$s which code elements of $X$, and don't attach these labels to the ones that don't. It's a good exercise to show that $A_X\cong A_Y\iff X=Y$, and so we do indeed get continuum-many pairwise-non-homeomorphic subspaces of $\mathbb{Q}$.
