Let $(K,v)$ be an algebraically closed complete valuation field. Let $(R,v|_R)$ be a valuation subring of $K$. Denote by $F^{\operatorname{sep}}$ the separable closure of $\operatorname{Frac}R$ in $K$.
If $\operatorname{Frac}R$ is complete w.r.t. $v|_{\operatorname{Frac}R}$, then the standard algebraic number theory argument tells us that the valuation $v|_{\operatorname{Frac}R}$ on $\operatorname{Frac}R$ extends uniquely to $F^{\operatorname{sep}}$ (which, obviously coincides with $v|_{F^{\operatorname{sep}}}$ ) and thus $v|_{F^{\operatorname{sep}}}\circ g=v|_{F^{\operatorname{sep}}} $ for every $g\in\operatorname{Gal}(F^{\operatorname{sep}}/\operatorname{Frac}R)$.
Now the question is: what if we no longer assume that $(\operatorname{Frac}R,v|_{\operatorname{Frac}R})$ is complete? Do we still have $v|_{F^{\operatorname{sep}}}\circ g=v|_{F^{\operatorname{sep}}} $ for every $g\in\operatorname{Gal}(F^{\operatorname{sep}}/\operatorname{Frac}R)$?
In this situation, it is unclear whether the extension of the valuation is unique, therefore we can't use the "norm construction" for $v|_{F^{\operatorname{sep}}}$.
P.S. No valuation here is assumed to be discrete.
Absolutely not.
Let $(K,v) = (\mathbf C_p,|\cdot|_p)$ and $R = \mathbf Z_{(p)}$, so ${\rm Frac}(R) = \mathbf Q$. You are asking, for each $g \in {\rm Gal}(\overline{\mathbf Q}/\mathbf Q)$, if we have $|g(\gamma)|_p = |\gamma|_p$ for each $\gamma$ in $\overline{\mathbf Q}$.
Assume the answer to your question is yes, and let $f(x)$ be a polynomial in $\mathbf Q[x]$ of degree $n$ that is irreducible over $\mathbf Q$ but is reducible over $\mathbf Q_p$. Let $\alpha$ be a root of $f(x)$ in $\overline{\mathbf Q} \subset \mathbf C_p$ and let $d \geq 1$ be the degree of its minimal polynomial over $\mathbf Q_p$. Then the constant term of that minimal polynomial over $\mathbf Q_p$ has absolute value $|\alpha|_p^d$. Every root of $f(x)$ in $\mathbf C_p$ has the form $g(\alpha)$ for some $g \in {\rm Gal}(\overline{\mathbf Q}/\mathbf Q)$, and the constant term of the minimal polynomial of $g(\alpha)$ over $\mathbf Q_p$ has absolute value $|g(\alpha)|_p^{d'} = |\alpha|_p^{d'}$ for some $d' \geq 1$.
If $|\alpha|_p^d < 1$ then $|\alpha|_p < 1$, so $|g(\alpha)|_p^{d'} < 1$. So if some irreducible factor of $f(x)$ over $\mathbf Q_p$ has a constant term with $p$-adic absolute value less than $1$ then this property is true for the constant terms of all irreducible factors of $f(x)$ over $\mathbf Q_p$. But there are lots of counterexamples to this. That is, there are lots of monic irreducible polynomials in $\mathbf Q[x]$ that have some monic irreducible factor in $\mathbf Q_p[x]$ with a constant term in $p\mathbf Z_p$ and another monic irreducible factor in $\mathbf Q_p[x]$ with a constant term in $\mathbf Z_p^\times$
Example. The polynomial $x^2 +2x+5$ is irreducible over $\mathbf Q$ but reducible over $\mathbf Q_5$ with irreducible factors $x+r$ and $x+s$ where $r = 3 \cdot 5 + 4 \cdot 5^2 + \cdots$ is in $5\mathbf Z_5$ and $s = 2 + 2 \cdot 5 + \cdots$ is in $\mathbf Z_5^\times$.
Example. The polynomial $x^3 +x+7$ is irreducible over $\mathbf Q$ but reducible over $\mathbf Q_7$ with irreducible factors $x+r$ and $x^2+sx+t$ where $r = 7 + 6 \cdot 7^3 + \cdots$ is in $7\mathbf Z_7$ and $t = 1 + 7^2 + \cdots$ is in $\mathbf Z_7^\times$.