Let $(M,\alpha)$ be a contact manifold. It is well-known that the critical points of the action functional, given by
\begin{align} A: C^\infty(S^1,M) \to \mathbb{R} \\ \gamma \mapsto \int_\gamma \alpha \end{align} are the closed trajectories of the Reeb flow. I tried to do the computation but failed.
Take a one-parameter family of loops $\gamma_t: S^1 \to \mathbb{R}$ such that $\gamma_0=\gamma$. Now compute $$\frac{d}{dt}|_{t=0}A(\gamma_t)= \frac{d}{dt} |_{t=0}\int_{S^1} \gamma_t^*\alpha= \int_{S^1}\gamma^*\mathcal{L}_{X}\alpha$$ where $X=\frac{d}{dt}|_{t=0}\gamma_t$. By Cartan formula, this equals $$\int_{S^1}\gamma^* (di_{X}\alpha+i_{X} d\alpha).$$ Being a critical point means that this quantity is zero. How can I conclude that $\gamma$ is a closed trajectory of the Reeb flow? For this, I would need that $\alpha(\dot{\gamma})=constant$ and $d\alpha (\dot{\gamma},\cdot)=0$.
There are some issues regarding that Lie derivative (see here for more information. It is a closely related question).
But let's proceed as if there isn't any issue.
We have \begin{align*}\require{cancel} \int_{S^1}\gamma^* (di_{X}\alpha+i_{X} d\alpha) &=\cancelto{0, \text{ by Stokes}}{\int_{S^1}d(\gamma^*\iota_X\alpha)}+\int_{S^1}\gamma^*\iota_Xd\alpha \\ &=\int_0^1d\alpha(X,\dot{\gamma}). \end{align*} This is zero for every $X$ if and only if $d\alpha(\cdot,\dot{\gamma})$ is zero. Which gets what you wanted...
... well, not everything. You wanted the orbit to be such that $\alpha(\dot{\gamma})=cnst$. You can't infer that, though. What you can infer is only that $\gamma$ is a reparametrization of the Reeb flow (not necessarily of constant speed), since the derivative always lies in the same $1$-distribution of $TM$ (which is $\ker d\alpha$).