Let $R$ be a henselian ring. Is $R[[x]]$ also a henselian ring?
Yes, it is. More generally, the following result holds.
Proposition. If $R$ is henselian at the maximal ideal $\mathfrak{m}$, then $R[[x_1, \ldots, x_n]]$ is henselian at the maximal ideal lying over $\mathfrak{m}$.
A reference is the paper by N. Sankharan A Theorem on Henselian Rings, Canad. Math. Bull. 11 (1968), 275277. See in particular Corollary 2.
Remark. The Proposition above is no longer valid if one takes the polynomial ring instead of the power series ring. For instance, if $K$ is a field than $K$ is henselian but $K[x]$ is not.

9$\begingroup$ More generally: if $A$ is a local ring and $I$ an ideal such that $A$ is $I$adically complete and separated, then $A$ is henselian if (and only if!) $A/I$ is henselian. This is easy by observing that each
$A_n:=A/I^n$
is henselian (because $A_{n,\mathrm{red}}$ is) and then inductively lifting roots to $A$. See Raynaud, LNM 169, I, §2 where this result is given as an exercise. $\endgroup$ – Laurent MoretBailly Feb 18 '13 at 14:46