To complement Hailong's answer, here is another way of seeing that an irreducible component of a Cohen-Macaulay scheme need not have special properties. In the example below, the whole scheme is Cohen-Macaulay, so all of its components have the same dimension, unlike in Hailong's example.

Let $X$ be a reduced and irreducible subscheme of projective space $\mathbb{P}^n$ of codimension $c$. Choose $c$ elements $f_1,\ldots,f_c$ in the ideal of $X$ in $\mathbb{P}^n$ with the property that the vanishing set $X'$ of $f_1,\ldots,f_c$ is reduced and has codimension $c$. Thus the scheme $X'$ contains $X$ as a component, it is a complete intersection, and hence it is Cohen-Macaulay. On the other hand, $X$ was essentially arbitrary, so you could have chosen it to be not Cohen-Macaulay!

For a more explicit example, let $X$ be a surface in $\mathbb{P}^4$ with a point that analytically locally looks like the union of two planes at a single point. (This is one of the standard example of a scheme that is not Cohen-Macaulay.) Choose two "general" elements of the ideal of $X$, and let $X'$ be the scheme defined by those two elements. Thus, $X'$ is Cohen-Macaulay and $X$ is a component of $X'$, but the surface $X$ it is not Cohen-Macaulay.

**EDIT** As Hailong pointed out, I answered a question that is different than what was asked. To answer the initial question, it suffices to argue as above, choosing also a hypersurface $H$ in $\mathbb{P}^n$ not containing $X$. Denote by $h$ an equation for $H$ and replace $f_1 , \ldots , f_c$ by $h f_1 , \ldots , h f_c$. The vanishing set $\overline{X}$ of these equations consists of the union of $H$ and the scheme $X'$ we had before. Thus $X$ is still a component of $\overline{X}$, of codimension $c$ in $\mathbb{P}^n$, the ideal of $\overline{X}$ is generated by $c$ equations, but the component $X$ is (essentially) arbitrary, in particular it need not be Cohen-Macaulay. This should now answer the question that was asked!