Nik Weaver has already explained in his answer that this does not hold, in general. On the positive side, here is a sufficient condition for the implication to be true:
Proposition. Let $\Omega_1, \Omega_2$ be topological spaces and let $T: C_b(\Omega_1) \to C_b(\Omega_2)$ be a positive linear operator such that $T1 = 1$. Suppose in addition that $T$ has the following continuity property:
$(*)$ If a sequence $(g_n) \subseteq C_b(\Omega_1)$ is bounded in supremum norm and converges pointwise to $g \in C_b(\Omega_1)$, then $(Tg_n)$ converges pointwise to $Tg$.
Then $Tf$ has no zeros whenever $0 \le f \in C_b(\Omega_1)$ has no zeros.
Proof. Assume that $0 \le f \in C_b(\Omega_1)$ has no zeros. Then $(nf) \land 1$ converges pointwise to $1$ as $n \to \infty$. Hence, $T\big((nf) \land 1\big)$ converges pointwise to $T1 = 1$ as $n \to \infty$. But we have
nTf \ge T\big((nf) \land 1\big)
for each $n$, so $Tf$ cannot be $0$ at any point of $\Omega_2$. qed
Remark 1. In the statement (and proof) of the proposition, the function $1$ can be replaced with any other function $0 \le h \in C_b(\Omega_1)$ that does not have any zeros.
Remark 2. The continuity condition $(*)$ is much more common than one might, maybe, expect at first glance: it is satisfied for all transition operators that are given by measurable transition kernels (a class of operators which occurs frequently in stochastic analysis).