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 non-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 by $0$ 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).

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).