No : just take $A = \mathbb{F}_p$.
This is the only counterexample : any complete local sub-algebra $\iota_A \colon A \hookrightarrow {\Bbb F}_p[[X]]$, with $A \neq \mathbb{F}_p$, is noetherian of Krull dimension 1.
Indeed, if $A \neq \mathbb{F}_p$, then $\mathfrak{m}_A \neq 0$ (if not, then $A$ would be a field, and the existence of the morphism $A \rightarrow {\Bbb F}_p[[X]] \rightarrow \mathbb{F}_p$ would imply $A = \mathbb{F}_p$). Then :
$\mathfrak{m}_A {\Bbb F}_p[[X]]$ is a nonzero ideal of ${\Bbb F}_p[[X]]$, so that ${\Bbb F}_p[[X]] / \mathfrak{m}_A {\Bbb F}_p[[X]]$ is a finite dimensional $\mathbb{F}_p$-vector space.
$\mathfrak{m}_A {\Bbb F}_p[[X]]$ is contained in $(X)$ (in particular, the $A$-module ${\Bbb F}_p[[X]]$ is $\mathfrak{m}_A$-adically complete, since it is $X$-adically complete). If it wasn't, then we would have some equation
$$ 1 = \sum_i a_i f_i, $$
with $a_i \in \mathfrak{m}_A$ and $f_i \in {\Bbb F}_p[[X]]$. The element $1 - \sum_i a_i f_i(0)$ of $A$ would then be an invertible element of $A$, and thus of ${\Bbb F}_p[[X]]$, which belongs to $(X)$ ; a contradiction.
As noticed by the OP in his comment, these two facts imply that $\iota_A$ is finite. Now, ${\Bbb F}_p[[X]]$ has Krull dimension $1$, and $\iota_A$ is integral and injective, so that $A$ has Krull dimension $1$ by Cohen-Seidenberg theorem. Moreover, $A$ is noetherian by Eakin's theorem (Eakin, P.M. "The Converse to a Well Known Theorem on Noetherian Rings." Mathematische Annalen 177 (1968): 278-282).