The dilaton $\sigma$ is the Goldstone boson of scale invariance. Scale transformations $x\rightarrow x/\lambda$ are generated non linearly, e.g. $$ \sigma(x)\rightarrow \sigma(\lambda x)+f \log\lambda\,,\qquad \lambda>0 $$ where $f$ is the dilaton decay constant (see below). An effective field theory for this Goldstone boson can be easily written with the following trick described e.g. in this very nice paper. In particular, define $\chi(x)=f e^{\sigma(x)/f}$ so that under scale transformation $\chi$ behaves like a field of scaling dimension $1$ $$ \chi(x)\rightarrow \lambda\, \chi(\lambda x)\,. $$ With this field at hand now you can build your scale invariant lagrangian that describes the effective theory of the dilaton: it is an expansion in derivatives $\mathcal{L}(\chi)= \sum_{n} \mathcal{L}^{(2n)}$ where the lagrangian $\mathcal{L}^{(n)}$ is \begin{equation} \mathcal{L}^{(2n)}=\sum_{m\geq 0} \frac{a_{n,m}}{(4\pi)^{2n-2}f^{2n-4}}\frac{\partial^{2n}\chi^m}{\chi^{2n+m-4}} \end{equation} where the derivatives are inserted in all possible ways among the $\chi^m$. 976
The logic is to write all operators with scaling dimension 4 so that the action $S=\int d^4 x \mathcal{L}$ will be scale invariant (given that $d^4 x\rightarrow \lambda^{-4}d^4 x$). You can basically think of $\chi$ as conformal compensator: build a Lorentz invariant and divide it by enough powers of $\chi$ to make of dimension 4. Notice that a term with no derivatives, contrary the the usual Goldstone bosons of internal symmetries, is instead allowed: $\mathcal{L}^{(4)}=(4\pi)^2a_{0,0}\chi^4$.
Moreover, if the dilaton isn't the only light state in the effective theory left over by the spontaneous symmetry breaking, other terms that seem breaking scale invariance need to be compensated by the insertions of the Goldstone so that dynamics will be invariant (whereas the states of the theory will not). For example, since you have mentioned QCD, one could wonder that the RG evolution comes from spontaneous breaking of scale invariance. Some particles that were contributing to the beta functions (which was zero in an unbroken scale invariant theory) becomes massive after spontaneous symmetry breaking, then decouple and don't contribute to $\beta$ anymore, leaving in fact an imbalance for which $\beta$ start running from the scale of symmetry breaking down to smaller energy scales. In practice one knows that the coupling runs logaritmically from the scale $f$ of symmetry breaking down to an arbitrary IR scale $\mu_{IR}$ $$ \mathcal{L}{gauge}=-\frac{F^2{\mu\nu}}{4g^2(\mu_{IR})}=-\frac{F_{\mu\nu}^2}{4}\left[\frac{1}{g^2(\mu_{UV})}+\frac{b_{IR}}{(4\pi)^2}\log\frac{f}{\mu_{IR}}\right] $$ But the scale $f$ appeared spontaneously and needs then to be compensated by dilaton insertion to restore scale invariance, $f\rightarrow f e^{\sigma(x)/f}=\chi(x)$, so that one can immediately read off the coupling among the gauge bosons and the dilaton \begin{align} \mathcal{L}{gauge} & \rightarrow -\frac{F{\mu\nu}^2}{4}\left[\frac{1}{g^2(\mu_{UV})}+\frac{b_{IR}}{(4\pi)^2}\log\frac{f e^{\sigma(x)/f}}{\mu_{IR}}\right]\ =& \mathcal{L}{gauge}-\frac{b{IR}}{(4\pi)^2 f}\sigma(x)F_{\mu\nu}^2\,. \end{align} As you can see the dilaton couples to the the $\beta$ function. 976