With your particular choice of $R$, the $H^2$ is $0$. More generally, if $R$ is a strictly Henselian *regular* local ring of dimension $2$, then by the purity for the Brauer group (in this particular case it is known and due to, I believe, Grothendieck; for a proof see Grothendieck "Le groupe de Brauer II", Prop. 2.3) $H^2_{et}(R \setminus \{ \mathfrak{m} \}, \mathbf{G}_m) = H^2_{et}(R, \mathbf{G}_m) = 0$. The first equality is due to the purity because you're removing a closed subscheme of codimension $2$; the second equality is because $R$ is strictly Henselian and $\mathbf{G}_m$ is smooth: by Grothendieck "Le groupe de Brauer III" appendix, Thm. 11.7 2), the cohomology can thus be computed over the separably closed residue field, where it vanishes.

More generally assuming that $R$ is a strictly Henselian regular local ring of dimension $\ge 2$, the $H^2$ is always supposed to be $0$ by the aforementioned purity and the argument as above. This is open (as far as I know), although many cases are known, including $\dim R \le 3$. For a brief survey on precisely this question see Gabber "On purity for the Brauer group" in Oberwolfach report No. 37/2004.