Is the kth central moment less than the kth raw moment for even k?

The statement is not true in general and it is easy to construct a counterexample. For example, we want $\mathbb{E}(X^4)\leq \mathbb{E}((X-\mathbb{E}(X))^4)$, we can let $X$ has a small probability of taking a large positive value, while keeping $\mathbb{E}(X)$ negative. Let $X$ be a random variable with $P(X=-2)=1-\epsilon$ and $P(x=M)=\epsilon$. We choose $\epsilon=1/(M+2)$ such that $\mathbb{E}(X)=-1$.

Then we have $$ \mathbb{E}(X^2)=M+2,\quad \mathbb{E}((X-\mathbb{E}(X))^2)=M+1 $$ and $$ \mathbb{E}(X^4)=M^3-2M^2+4M+8,\quad \mathbb{E}((X-\mathbb{E}(X))^4)=M^3+2M^2+2M+1. $$ Obviously, if $M$ is large enough, we have $\mathbb{E}(X^4)\leq \mathbb{E}((X-\mathbb{E}(X))^4)$.

To complement the accepted answer, we can show that $E[(X - E[X])^m] \leq E[X^m]$ when $X \geq 0$. Assume without loss of generality that $E[X] = 1$ (rescaling $X$ if necessary). Then, omitting the dependence on $\omega$ for notational convenience, \begin{align} & \int_{\Omega} X^m \, d \Omega - \int_{\Omega} (X - 1)^m \, dP(\omega) \\ & \quad = \int_{X \leq 1/2} \left[ X^m - (X - 1)^m \right] \, dP(\omega) + \int_{1/2 < X \leq 1} \left[ X^m - (X - 1)^m \right] \, dP(\omega) \\ & \quad + \int_{X > 1} \left[ X^m - (X - 1)^m \right] \, dP(\omega). \\ & \quad \geq \int_{X \leq 1/2} \left[ X^m - (X - 1)^m \right] \, dP(\omega) + \int_{X > 1} \left[ X^m - (X - 1)^m \right] \, dP(\omega) =: I_1 + I_2. \end{align} For $m$ odd, both $I_1$ and $I_2$ are positive and the statement is proved. We assume from now on that $m$ is even. It is easy to see that: \begin{align} |a^m - (a - 1)^m| \leq (1 - a) \quad & \text{ if } \quad 0 \leq a \leq 1/2; \\ a^m - (a - 1)^m \geq (a - 1) \quad & \text{ if } \quad a > 1. \end{align} Using these inequalities, we deduce that $$ I_1 + I_2 \geq \int_{X > 1} (X - 1) \, dP(\omega) - \int_{X \leq 1/2} (1 - X) \, dP(\omega). $$ To conclude, we note that, by definition of the expected value, \begin{align} & \int_{X \leq 1/2} (X(\omega) - 1) \, dP(\omega) + \int_{1/2 < X \leq 1} (X(\omega) - 1) \, dP(\omega) + \int_{X > 1} (X(\omega) - 1) \, dP(\omega) = 0 \\ & \rightarrow \int_{X > 1} (X(\omega) - 1) \, dP(\omega) \geq \int_{X \leq 1/2} (1 - X(\omega)) \, dP(\omega). \end{align}

EDIT: Ixob Nil's proof in a different answer is a nice simplification of this proof.

Roberto Rastapopoulos' proof could be simplified by firstly proving the inequality holds when $a\ge 0$: $$a^m-(a-1)^m\ge a-1;$$

Then similarly for $X\ge0$ with $E(X)=1$, $$E(X^r)-E[(X-1)^r] = \int_{X\ge 0}[x^r-(x-1)^r]dP(\omega)\ge \int_{X\ge 0}(x-1)dP(\omega)=0.$$