Does the independency of $x_i|\theta$ implies the independency of $x_i$? Suppose we have a random sample $x_1, \cdots, x_n$ draw from the density $f(x|\theta)$ where $\theta$ is the parameter. If we know $x_i|\theta$ are independent of each other, i.e. $$f({\bf x}|\theta)=\prod^n_{i=1}f(x_i|\theta)$$ Does this implies $$f({\bf x})=\prod^n_{i=1}f(x_i)$$ where $f({\bf x})$ is marginals, we can get it from $$f({\bf x})=\int f({\bf x}, \theta)d\theta=\int f({\bf x}|\theta)\pi(\theta)d\theta$$ where $\pi(\theta)$ is some prior of $\theta$, and usually we can choose noninformative prior. I guess it is not...

Definitely not. Suppose that _conditioned on $\theta \in (0,1)$_ the $x_i$ are [conditionally] i.i.d. from $\text{Bernoulli}(\theta)$. Then $P(x_1=x_2=1) = E[\theta^2]$ while $P(x_1=1)P(x_2=1) = E[\theta]^2$. Thus $x_1$ and $x_2$ are marginally independent only if $\theta$ is a degenerate (constant) random variable.

Some intuition: the $x_i$ each have information about $\theta$, so they ought to be dependent on each other through $\theta$.