The answer is yes.
Proposition. Let $\alpha$ be the standard contact form on the three-sphere (for which the Reeb vector field is the Hopf vector field $X$). If $f$ is a strictly positive function on $S^3$ such there exist two Hopf circles $\gamma_1$ and $\gamma_2$ for which
$$
\int_{\gamma_1} f^{-1} \alpha \neq \int_{\gamma_2} f^{-1} \alpha \ ,
$$
then the vector field $fX$ is not the Reeb vector field of any contact form on $S^3$.
The key idea is that all solutions of $\dot{x}(t) = f(x(t))X(x(t))$ are periodic and that if $fX$ were a Reeb vector field, all these solutions would have the same period.
Indeed, assume $fX$ is the Reeb vector field of a contact form $\beta$. If $\gamma_1$ and $\gamma_2$ are integral curves of $fX$ with periods $T_1$ and $T_2$, then
$$
T_2 - T_1 = \int_{\gamma_2} \beta - \int_{\gamma_1} \beta = \int_\Sigma d\beta \ ,
$$
where $\Sigma$ is a cylinder with boundary $\gamma_2 - \gamma_1$ and foliated by Hopf circles (I'm being a bit sloppy with the word foliated). It follows that this last integral is zero and so $T_1 = T_2$.
Now we just have to notice that the period of $\gamma_i$ ($i = 1, 2$) equals
the integral of $f^{-1}\alpha$ over $\gamma_i$:
$$
\int_{\gamma_i} f^{-1} \alpha = \int_0^{T_i} f^{-1}(\gamma_i(t))\alpha(\dot{\gamma_i}(t)) \, dt = \int_0^{T_i} f^{-1}(\gamma_i(t))\alpha(f(\gamma_i(t)) X(\gamma_i(t))) \, dt = T_i
$$
