No. There is a counterexample for each $n\geq4$.

I will only show a counterexample for $n=4$; you can get one for any higher
$n$ by appending $0$'s.

Let $A_{1}=\left\{ \left( x,y,z,w\right) \in\left\{ 0,1\right\}
^{4}\ \mid\ x+y\leq1\text{ and }z+w\leq1\right\} $. Let $A_{2}=\left\{
0,1\right\} ^{4}\setminus A_{1}$. Then, $\left( A_{1},A_{2}\right) $ is
what you call an order-preserved partition (I have seen this being called an
"admissible partition"). Your conjecture claims that there exist $p_{1}
,p_{2},p_{3},p_{4}\in\left( 0,1\right) $ and $\theta\in\mathbb{R}_{+}$ such
that the function $f:\left\{ 0,1\right\} ^{4}\rightarrow\mathbb{R},\ \left(
x,y,z,w\right) \mapsto p_{1}^{1-x}p_{2}^{1-y}p_{3}^{1-z}p_{4}^{1-w}$
satisfies $A_{1}\subseteq f^{-1}\left( \left( -\infty,\theta\right)
\right) $ and $A_{2}\subseteq f^{-1}\left( \left( \theta,+\infty\right)
\right) $.

Assume that this were true. Then, there would exist $q_{1},q_{2},q_{3}
,q_{4}\in\left( 0,1\right) $ and $\eta\in\mathbb{R}$ such that the function
$g:\left\{ 0,1\right\} ^{4}\rightarrow\mathbb{R},\ \left( x,y,z,w\right)
\mapsto q_{1}x_{1}+q_{2}x_{2}+q_{3}x_{3}+q_{4}x_{4}$ satisfies $A_{1}\subseteq
g^{-1}\left( \left( \eta,\infty\right) \right) $ and $A_{2}\subseteq
g^{-1}\left( \left( -\infty,\eta\right) \right) $. (Indeed, take
$q_{i}=\ln p_{i}$ and $\eta=\ln p_{1}+\ln p_{2}+\ln p_{3}+\ln p_{4}-\ln\theta
$. Then, the inequalities $f\left( x,y,z,w\right) \lesseqgtr\theta$ become
$g\left( x,y,z,w\right) \gtreqless\eta$ after taking natural logarithms and
slightly rewriting.)

Now, the two points $\left( 1,0,1,0\right) $ and $\left( 0,1,0,1\right) $
both belong to $A_{1}$ and therefore to $g^{-1}\left( \left( \eta
,\infty\right) \right) $ (since $A_{1}\subseteq g^{-1}\left( \left(
\eta,\infty\right) \right) $). Since $g^{-1}\left( \left( \eta
,\infty\right) \right) $ is a convex set, the midpoint between these two
points (that is, the point $\left( \dfrac{1}{2},\dfrac{1}{2},\dfrac{1}
{2},\dfrac{1}{2}\right) $) must therefore also belong to $g^{-1}\left(
\left( \eta,\infty\right) \right) $.

But the two points $\left( 1,1,0,0\right) $ and $\left( 0,0,1,1\right) $
belong to $A_{2}$ and therefore to $g^{-1}\left( \left( -\infty,\eta\right)
\right) $ (since $A_{2}\subseteq g^{-1}\left( \left( -\infty,\eta\right)
\right) $). Since $g^{-1}\left( \left( -\infty,\eta\right) \right) $ is a
convex set, the midpoint between these two points (that is, the point $\left(
\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2}\right) $) must therefore
also belong to $g^{-1}\left( \left( -\infty,\eta\right) \right) $.

So the point $\left( \dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}
{2}\right) $ must belong to $g^{-1}\left( \left( \eta,\infty\right)
\right) $ and to $g^{-1}\left( \left( -\infty,\eta\right) \right) $ at
the same time. But this is absurd, since $g^{-1}\left( \left( \eta
,\infty\right) \right) \cap g^{-1}\left( \left( -\infty,\eta\right)
\right) =\varnothing$. So we have a contradiction.

I am wondering whether it is possible to fix your conjecture by (essentially)
assuming that the kind of counterexamples above (viz., two points in $A_{1}$
having the same midpoint as two points in $A_{2}$) does not exist.

A stronger requirement that definitely makes your conjecture true is the following:

**(1)** no convex combination of the points in $A_1$ can be a convex combination of the points in $A_2$ at the same time.

Indeed, if **(1)** is true, then the convex hull of $A_1$ is disjoint from the convex hull of $A_2$; but (from basic linear optimization theory) we know that this entails that there exists a hyperplane separating $A_1$ from $A_2$; this immediately translates into the existence of $g$ and $\eta$ as above.