The answer depends on the $2$-dimensional subbundle of $TM$.  

Suppose that $\Delta$ is a contact structure on $M^3$ with $\Delta$ defined by a $1$-form $\omega$ such that $\omega\wedge\mathrm{d}\omega\not=0$.  Let $X$ be the Reeb vector field for $\omega$, i.e., $\omega(X) = 1$ and $\iota(X)(\mathrm{d}\omega) = 0$.  (Here, $\iota(X)$ means the interior product with $X$.)

If $f$ is a nonzero function on $M$, and $\tilde\omega = f\omega$, then the Reeb vector field for $\tilde\omega$ is $\tilde X$, where
$$
\tilde X = \frac1{f}\ X + F
$$
where $F$ is the unique vector field that satisfies $\omega(F)=0$ and 
$$
\iota(F)(\mathrm{d}\omega) = \frac{\mathrm{d}f - \mathrm{d}f(X)\ \omega}{f^2}
$$

Suppose that you have a $2$-plane field on $M$ defined by the vanishing of a nonzero $1$-form $\theta$ on $M$.  Then you can write $\theta$ uniquely in the form $\theta = a\,\omega + \iota(B)(\mathrm{d}\omega)$ where $a$ is a smooth function on $M$ and $B$ is a vector field that satisfies $\omega(B) = 0$.  To know whether $\tilde X$ lies in the kernel of $\theta$, you just evaluate to get
$$
0 = \theta(\tilde X) = \frac{a}{f} + \iota(F)\bigl(\iota(B)(\mathrm{d}\omega)\bigr)
= \frac{a}{f} - \iota(B)\bigl(\iota(F)(\mathrm{d}\omega)\bigr) = \frac{a}{f}  - \frac{\mathrm{d}f(B)}{f^2}
$$
Thus, the condition on $f$ is the first order linear PDE $$\mathrm{d}f(B) - af = 0.$$
Since $f$ is nowhere vanishing, we can assume that $f$ is positive and write it in the form $f = e^u$ for some smooth function $u$ on $M$, so that the equation becomes
$$
\mathrm{d}u(B) = a.
$$
Note that, we can't have $B=0$ at any point where there is a solution because that would force $a$ to vanish there, which would make $\theta$ vanish there.  Thus, $B$ must be nowhere vanishing if there is to be a solution.  That's one obvious condition.

However, there are more global conditions:  For example, if $B$ has a closed integral curve and $a$ is positive along that integral curve, then there cannot be any solution $u$ to the above equation along this curve, because $u$ would have to be strictly increasing along this closed curve.  Since it is easy to construct a $B$ with a closed integral curve on such a $3$-manifold and then choose $a$ to be a positive function on $M$, there will always be $2$-plane fields for which there is no solution to your problem.

Another global condition comes directly from the formula for $\tilde X$.  Note that, at any critical point of $f$ (i.e., where $\mathrm{d}f$ vanishes), one must have $F$ vanish as well, in which case $\tilde X$ will be a multiple of $X$.  Thus, if your $2$-plane field $\theta=0$ is always transverse to the line field spanned by $X$ (i.e., if $a$ is nowhere vanishing), then there can't be a solution in this case either.