I have seen in a paper to use (this is my understanding, I may be wrong) the following for $p \geq 2$: $$ E[\|X_{n+1}-E[X_{n+1}|\sigma(X_n)]\|^p] \leq E[\|X_{n+1}\|^p] $$ where the norm is Euclidean norm and $\{X_n\}$ is a Markov process.
I sense Hilbert space projection idea here but that works only for $p =2$.
This is false, even for a trivial $\mathscr{F}_n$. I believe that this is false for each $p>2$, but will show for $p=4$.
Let $X = 1$ with probability $x$ and $-1$ with probability $1-x$. Then $E[X^4] = 1$, $$E[(X-E[X])^4] = x (2-2x)^4 + (1-x)(2x)^4.$$ E.g., for $x=1/4$, we have $$ E[(X-E[X])^4] = \frac{21}{16} > 1. $$
Here is a simple argument why this cannot be true for all $p\ge 2$. In fact, the inequality is equivalent to $||X-E[X|\mathscr{F}]||_p\le ||X||_p$, so by letting $p\to\infty$ we would get $\operatorname{ess\,sup} |X-E[X|\mathscr{F}]|\le \operatorname{ess\,sup} |X|$, which is definitely false. And my example uses this idea: you ensure that $\operatorname{ess\,sup} |X-E[X]|>\operatorname{ess\,sup} |X|$.