Strict stochastic domination of "thinned out" random cluster model Fix some $q\geq 1$ and denote by $X_p$ a random variable sampled from the law of the random cluster model with parameters $p,q$ on some graph $G$ and with, say, free boundary conditions. 
Define the "thinned out" $X_p^\delta$ as follows: Sample (an independent copy of) $X_p$ and then close each open edge independently with probability $\delta>0$. 
Equivalently one may define $X_p^\delta$ as the pointwise product of $X_p$ and an independent Bernoulli percolation process with parameter $1-\delta$.
Can one prove that $X_p^\delta$ is stochastically dominated by some $X_{p'}$ with $p'<p$? (Clearly it works for $\delta$ sufficiently large and $p'=p$, but I would like to say that one can find such a $p'=p'(\delta)<p$ for every $\delta>0$.)

My attempt: We have the finite energy property, i.e. no matter what values we condition on outside of some fixed edge $e$, the probability of that edge being open is in $\{p,p/(p+q(1-p))\}$; in particular it is bounded away from $1$ (and $0$). So by Lemma 1.1 we can stochastically bound the law of $X_p$ from above (and below) by some Bernoulli percolation measure. Clearly $X_p^\delta$ (as product of $X_p$ and some independent Bernoulli process) satisfies the same stochastic inequality with an even smaller bound, but I'm not sure if I can translate this back into stochastic domination in terms of a random cluster model. I guess I would need some (independent) process $Y$ which is stochastically bounded from below by some Bernoulli percolation process such that $YX_p$ (pointwise product) has the law of $X_{p'}$, but I don't know how to construct such a $Y$. 
Theorem 1.6 (MON) seems helpful.
 A: Because you stated that $G$ has free boundary conditions I assume that $G$ is finite.
In this case, for a fixed $\delta > 0$, it is easy to see that any increasing event $\mathcal{A}$ (except the whole set of events)
satisfies
$$\mu_p^\delta [\mathcal{A}] < \mu_p[\mathcal{A}]\,, $$
where $\mu_p$ is the law of $X_p$ and $\mu_p^\delta$ is the law of $X_p^\delta$. Therefore there will exist a $p'<p$ such that
$$\mu_p^\delta \leq \mu_{p'} \leq \mu_p $$
simply because for $p'\to p$, one has $\mu_{p'}[\mathcal{A}]\to\mu_{p}[\mathcal{A}]$ continuously (the set of increasing events $\mathcal{A}$ is finite). A formula for $p'$ derived from this argument will however severely depend on $G$.

My intuition tells me that one cannot find $p' = p'(p,q,\delta)$ which is independent of the graph. Consider the following 'counterexample':
We take $G=\mathbb{Z}^2$ to be the inifinite square lattice and set $q>4$, and $p = p_c = \frac{\sqrt{q}}{1+\sqrt{q}}$ the critical temperature. Let us consider the infinite volume measure $\mu_p$ which is obtained from the limit process on subgraphs of $\mathbb{Z}^2$ with wired boundary conditions. The model then is in the 'supercritical phase' (compare here), especially it should satisfy for an edge $e$
$$\mu_p[e \text{ is open}] > 0.5$$
(I did not check thoroughly, though i am pretty sure). Therefore,
after applying edge deletion with small enough $\delta$, one still has
$$\mu_p^\delta[e \text{ is open}] = (1-\delta)\mu_p[e \text{ is open}] > 0.5\,.$$
However, if we look at the infinite volume measure $\mu_{p'}$ for $p'<p = p_c$,
it will be in the subcritical phase, especially one gets
$$ \mu_{p'}[e \text{ is open}] < 0.5 $$
(this is easy to show using a duality argument).
From this it is clear that $\mu_{p}^\delta$ cannot be stochastically dominated by $\mu_{p'}$. Moreover, since the limit process to obtain the infinite volume measures is rather well-behaved (with respect to monotonicity properties), one can find large enough finite subgraphs $G$ of $\mathbb{Z}^2$ with wired boundary conditions such that the contradiction still follows.
