Let me first explain the statement of the question and then give some indication why the answer might be 'yes'.
By a space I mean, say, a simplicial set and by rational I mean rational in the sense of Bousfield, i.e. local with respect to the homology theory $H\mathbb{Q}$. If we denote the category of spaces by $\mathcal{S}$ then the category of rational spaces $\mathcal{S}\_\mathbb{Q} \subseteq \mathcal{S}$ is a full subcategory and the question is whether it is closed under filtered colimits. This would in particular imply that the $\infty$-category of rational spaces is compactly generated. Note that it is known that this $\infty$-category is generated by $\kappa$-compact objects for $\kappa$ the successor cardinal of $\omega$ (by results of Bousfield). But whether it is compactly generated is unclear to me and this is really what I want to know, even if the inclusion $\mathcal{S}\_\mathbb{Q} \subseteq \mathcal{S}$ is not necessarily $\omega$-continuous. 


Now let me give some indication why the answer could be 'yes': first a simply connected (or more generally nilpotent) space is rational if and only if its homotopy groups are uniquely divisible. This property is obviously closed under filtered colimits. Thus a filtered colimit of simply connected, rational spaces is again rational. Bousfield characterizes rationality for arbitrary spaces as a property of homotopy groups, but this property is also homotopy theoretic in nature so that I am unable to decide if it is preserved by filtered colimits.


The second indication is that one can study the stable analogue of my question: is a filtered colimit of rational spectra again rational. The answer is yes. This is in fact equivalent to the fact that rationalization is a smashing localization. 
This observation also shows that the class of $\mathbb{S}/p$-local spectra for a prime $p$ cannot be closed under filtered colimits since $p$-completion is not smashing. A consequence is that $\mathbb{F}\_p$-local spaces (a.k.a. $\mathbb{S}/p = M(\mathbb{Z}/p)$-local spaces) are not closed under filtered colimits. 



Is a filtered colimit of rational spaces again rational?
The answer to this, as a result of some discussions with Thomas, turns out to be no.


Consider the abelian groups $A\_n = \Bbb Q[x]/(x^n)$. Multiplication by $x$ embeds $A\_n$ into $A\_{n+1}$, and the direct limit of the sequence
$$
A\_0 \to A\_1 \to \dots
$$
is the group $A\_{\infty} \cong \Bbb Q[x^{\pm 1}] / \Bbb Q[x]$.


The group $\Bbb Q$ acts compatibly on the $A\_n$ and $A\_\infty$ if we define
$$
q \ast m = (1+x)^q m
$$
where $(1 + x)^q = \sum \binom{q}{n} x^n$ is defined using the binomial theorem. 


The action of $\Bbb Q$ on each $A\_n$ is *nilpotent*: there is a finite filtration whose layers are acted on trivially. In fact, $\Bbb Q$ acts trivially on the quotient groups $A\_n/A\_{n-1} \cong \Bbb Q$ because $x$ acts trivially on this group.


By contrast, the action of $\Bbb Q$ on $A\_\infty$ is *perfect*: for any element $m \in A\_\infty$, we find
$$
m = x (x^{-1} m) = 1 \ast (x^{-1} m) - (x^{-1} m)
$$
and so the set of elements of the form $q \ast y - y$ generate all of $A\_\infty$. Therefore, $A\_\infty / \Bbb Q = 0$.


There is an associated sequence of Eilenberg--Mac Lane spaces $K(A\_n,2)$ and $K(A\_\infty,2)$ with compatible actions of $\Bbb Q$, for example by using a functorial construction of E-M spaces. We can form the Borel constructions $X\_\alpha = K(A\_\alpha,2) \times\_{\Bbb Q} E\Bbb Q$. These spaces have the property that
$$
\pi\_n X\_\alpha = \begin{cases}
 \Bbb Q &\text{if }\*=1,\\
 A\_\alpha &\text{if }\*=2,\\
 0 & \text{otherwise}.
\end{cases}
$$
In particular, $X\_\infty$ is the homotopy colimit of the spaces $X\_n$.


Each $X\_n$ is a *nilpotent space*: it has abelian (hence nilpotent) fundamental group, the fundamental group acts nilpotently on the higher homotopy groups, and the homotopy groups are rational vector spaces. This means that $X\_n$ has a Postnikov-type construction formed by iteratively taking fibers of maps to $K(\Bbb Q,m)$, and so $X\_n$ is local with respect to rational homology.


By contrast, $X\_\infty$ is not local with respect to rational homology. The fact that $\pi\_1(X\_\infty)$ acts perfectly on $\pi\_2(X\_\infty)$ allows us to use an analogue of Quillen's plus-construction to find a map $X\_\infty \to Y$, by attaching $3$-dimensional and $4$-dimensional cells, which is a rational homology isomorphism but which induces the zero map $\pi\_2(X\_\infty) \to \pi\_2(Y)$. In particular, the map $X\_\infty \to (X\_\infty)\_{\Bbb Q}$ to its rationalization factors through $Y$, and so it must send all of $\pi\_2$ to zero.


