= Solution
On the coordinatewise ordered <configuration space>, $\mu_1$ stochastically dominates $\mu_2$ when $\int f\,d\mu_1\geq\int f\,d\mu_2$ for every increasing real function $f$. Equivalently, $\mu_1(A)\geq\mu_2(A)$ for every <increasing event>. The <Holley condition> is the sufficient cross-lattice inequality
$$
\boxed{\mu_1(\omega\vee\eta)\mu_2(\omega\wedge\eta)
\geq\mu_1(\omega)\mu_2(\eta)\quad\text{for all }\omega,\eta.}
$$
Here join and meet are coordinatewise maximum and minimum. This condition implies $\mu_1\geq_{\rm st}\mu_2$ by the <Holley inequality>; it is not asserted to be necessary. It can be checked on unnormalized weights, since their normalization constants cancel.
Use the weights
$$
b_p(\omega)=p^{o(\omega)}(1-p)^{|E|-o(\omega)},\qquad
r_{p,q}(\omega)=b_p(\omega)q^{k(\omega)}.
$$
For the upper bound, take $\mu_1=\phi_{p,1}$ and $\mu_2=\phi_{p,q}$. Since open-<edge> count is modular, the <Bernoulli> factors cancel in the Holley ratio, leaving
$$
\frac{b_p(\omega\vee\eta)r_{p,q}(\omega\wedge\eta)}
{b_p(\omega)r_{p,q}(\eta)}
=q^{k(\omega\wedge\eta)-k(\eta)}\geq1.
$$
The last inequality uses $q\geq1$ and the fact that removing <edges> cannot reduce the number of components. Thus $\phi_{p,q}\leq_{\rm st}\phi_{p,1}$.
For the lower bound, let $p'=p/[p+q(1-p)]$. Its odds satisfy
$$
\frac{p'}{1-p'}=\frac1q\frac p{1-p}.
$$
Set $a=o(\omega\vee\eta)-o(\omega)=o(\eta)-o(\omega\wedge\eta)$, the number of <edges> added to $\omega$ in the union. For $\mu_1=\phi_{p,q}$ and $\mu_2=\phi_{p',1}$, the corresponding ratio is
$$
\frac{r_{p,q}(\omega\vee\eta)b_{p'}(\omega\wedge\eta)}
{r_{p,q}(\omega)b_{p'}(\eta)}
=q^{a+k(\omega\vee\eta)-k(\omega)}\geq1.
$$
Adding one <edge> either merges two components or leaves the count unchanged. Therefore adding $a$ <edges> reduces $k$ by at most $a$, proving the exponent nonnegative. This also proves the hinted <monotonicity> of $k(\omega)+o(\omega)$ rather than assuming it. The <Holley condition> now proves both <Bernoulli bounds for the random-cluster model>:
$$
\boxed{\phi_{p',1}\leq_{\rm st}\phi_{p,q}\leq_{\rm st}\phi_{p,1}}.
$$
The same argument holds with wired boundary conditions: identify the wired boundary vertices first and count components on that quotient graph. Taking finite-volume wired limits preserves every increasing cylinder-event inequality. Approximating $\{0\leftrightarrow\infty\}$ by the decreasing events $\{0\leftrightarrow\partial[-n,n]^d\}$ consequently gives
$$
\theta(p',1)\leq\theta^{\rm w}(p,q)\leq\theta(p,1).
$$
Let $a=p_c(1)$. For $p<a$, the upper bound makes the wired <percolation> <probability> zero, so $p_c(q)\geq a$. If $a<1$ and $p'>a$, the lower bound makes it positive. Solving this strict inequality,
$$
\frac p{p+q(1-p)}>a
\quad\Longleftrightarrow\quad
p>\frac{qa}{1+(q-1)a}.
$$
Taking infima proves
$$
\boxed{p_c(1)\leq p_c(q)\leq
\frac{q\,p_c(1)}{1+(q-1)p_c(1)}}.
$$
For $a=1$, the lower bound and the trivial upper bound $p_c(q)\leq1$ give the same conclusion. No assumption about <percolation> at the critical parameter is required.
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