Consider $\mathbb{R}^3$ (or more generally $\mathbb{R}^{2n-1}$), are there two different contact forms $\alpha_1 \neq \alpha_2$ with the same Reeb vector field and exterior derivative i.e. $d\alpha_1=d\alpha_2$?
Equivalently, does Reeb vector field and $d\alpha$ determine $\alpha$?
No, this data doesn't determine $\alpha$. Here's a counterexample.
On $\mathbb R^3$, consider \begin{align*} \alpha_1 &= x\,dy + dz,\\ \alpha_2 &= x\,dy + dz + dx. \end{align*} Then $d\alpha_1 = d\alpha_2 = dx\wedge dy$, and both Reeb fields are equal to $\partial/\partial z$.
More generally, if $\alpha_1$ is any contact form whatsoever and $T_1$ is its Reeb field, we can take $\alpha_2 = \alpha_1 + du$, where $u$ is any smooth function such that $T_1(u) \equiv 0$.
(It's always possible to find a nontrivial such example locally. Whether it can be done globally is a more subtle question that I don't know a general answer to. For example, if $T_1$ has a dense orbit, then every solution to $T_1(u)=0$ is constant.)