The statement is correct and here is one way to prove it. The slogan is that "finite etale extensions of $k(x)$ and $\widehat{k(x)}$ are the same for "quasi-complete" fields $k(x)$" (this is a toy example of Gabber's Approximation Lemma).
I decided to write a proof in complete details mostly to be sure that there is no small mistake (there are always subtle points when trying to work with general (even strongly noetherian) adic spaces).
We start with a formulation of a very important Lemma (whose proof I will explain at the end of this answer)
Lemma 1: Let $X:=\operatorname{Spa}(B, B^+)$ and $Y:=\operatorname{Spa}(A,A^+)$ be strongly noetherian affinoid adic spaces, $f: X\to Y$ be a finite morphism. Choose any point $y\in Y$ and denote its preimage (as a finite set of points) by $\{x_i\}$. Then we have an isomorphism $\mathcal O_{Y, y}\otimes_A B = \prod_{i} \mathcal O_{X,x_i}$.
Let us now use this Lemma in the case of interest. Namely, we have finite etale morphism $f:X \to Y$ of strongly noetherian affinoid adic spaces. Then Lemma 1 implies that an $A$-homorphism $\mathcal O_{Y,y} \to \prod_i \mathcal O_{X, x_i}$ is a finite etale morphism! In particular, an extension of a maximal ideal $\mathfrak{m}_y$ to the ring $\mathcal O_{X,x_i}$ is equal to $\mathfrak m_{x_i}$ for all $i$ (i.e. $\mathfrak{m}_y \mathcal O_{X,x_i}=\mathfrak{m}_{x_i}$) and each residue field $k(x_i)$ is a finite separable extension of $k(y)$. In other words, $$ k(y)=\prod_i k(x_i). $$ and each $k(x_i)$ is a finite separable extension of $k(y)$. Once we know that at least one of $k(x_i)$ is equal to $k(y)$ we can leverage the fact that $\mathcal O_{Y,y}$ is henselian to get a section analytically locally on $Y$ (this is well-explained in your question).
Proposition 2: Under notation as above, assume that $f:X \to Y$ has a section $$s: \operatorname{Spa}(k(y), k(y)^+) \to X,$$ then there is $i$ s.t. $k(x_i)=k(y)$.
Consider "a fiber" $X_y$, which is defined as a fiber product $X\times_Y \operatorname{Spa}(\widehat{k(y)}, \widehat{k(y)^+})$, this is a finite etale scheme over "a point" $\operatorname{Spa}(\widehat{k(y)}, \widehat{k(y)^+})=\operatorname{Spa}(k(y), k(y)^+)$ (this space is not necessary a one point space). Since finiteness is preserved by base change (at least for adic morphisms), we know that $X_y$ is a finite adic space over $\operatorname{Spa}(\widehat{k(y)}, \widehat{k(y)^+})$ and its algebra of functions is equal to $$ \mathcal O_{X_y}(X_y)=(k(y)\otimes_{A} B)\hat{}= (k(y)\otimes_{\mathcal O_{Y,y}} \mathcal O_{Y,y} \otimes_A B)\hat{}=$$ $$=(k(y)\otimes_{\mathcal O_{Y,y}} \prod_i \mathcal O_{X,x_i})\hat{}=(\prod_{i} k(x_i))\hat{}=\prod_i \widehat{k(x_i)}. $$ Existence of a section $s:\operatorname{Spa}(\widehat{k(y)},\widehat{k(y)^+}) \to X_y$ implies that there is an integer $i$, s.t. $\widehat{k(x_i)}=\widehat{k(y)}$. Now we want to use some Approximation result to conclude that $k(x_i)=k(y)$. One way to show this is to prove that a natural functor $$ \{\text{finite separable extensions of } k(y) \} \to \{\text{finite separable extensions of} \widehat{k(y) } \} $$ defined by $$ l \mapsto \widehat{l} $$ is an equivalence of categories.
This would mean that an isomorphism $\widehat{k(x_i)}\cong \widehat{k(y)}$ implies an existence of an isomorphism $k(x_i) \cong k(y)$. It is possible to prove this statement using a (very) difficult theorem of Gabber, but there is a way to avoid it.
Indeed, we know that $A$ is an analytic ring, hence any valuation has a unique vertical generalization. This means that the field $k(y)$ has a rank-1 valuation subring $k(y)^{\circ}$. Since we want to relate finite separable extensions of $k(y)$ to those of $\widehat{k(y)}$, we can forget about $k(y)^+$ and work with its rk-1 valuation instead. Then there is a theorem of Berkovich (Thm 2.4.1 here) that shows the claim for quasicomplete fields and there is another Proposition in the same paper (Proposition 2.4.3) that says to prove quasi-completeness of a rk-1 valutation field $k(y)$ it is sufficient to show that $k(y)^{\circ}$ is henselian (with respect to an ideal generated by a pseudo-uniformizer). But this is a standard fact for analytic Huber pairs.
All in all, we have the desired equivalence of categories, thus we conclude that $k(x_i)=k(y)$ and this is exactly what we were looking for.
Proof of Lemma 1: The answer turned out to be bigger than I expected, so I only sketch a proof of this Lemma. If somebody is interested in a complete proof, I can add it later.
The Lemma basically follows from the following four facts.
If two points $x$ and $y$ of a spectral space (such as $\operatorname{Spa}(A, A^+)$) don't have a common generalizing point, then there exist disjoint open neighborhoods containing $x$ and $y$.
In a notation as above none of the points $x_i$ has a common generalizing point. Hence, there are disjoint open neighbourhoods $U_i$ of each $x_i$.
Any finite morphism is closed, then a standard trick allows us to choose an open subset $U \subset \cup U_i$ of the form $U=f^{-1}(V)$ for some open $V$ containing $y$.
The last bullet point implies (with a little work) that $\mathcal O_{Y,y}\otimes_A B = \prod_i \mathcal O_{X, x_i}$.