This might not be a question appropriate for this forum, I apologize in this case... Is it true that any element of the completion of a valued ring $R$ that is algebraic over the field of fractions of $R$ is in the henselization of $R$ ?

Edit Add noetherian hypothesis because $R$ might no be a subring of $\hat{R}$ otherwise. The answer is no as pointed out by quasicoherent in the comments. But suppose $R$ is a discrete valuation ring and denote $K=\mathrm{Frac}(R)$, then
There are several equivalent properties defining henselien local rings. I will use two of them: let $(A, m)$ be a local ring and $k=A/m$. Then $A$ is henselian iff (a) for any monic polynomial $f(X)\in A[X]$, any simple root of $\bar{f}(X)\in k[X]$ lifts to root of $f(X)$ in $A$. or equivalently (b) If $A\to A'$ is a local homomorphism, étale and with trivial residue extension, then $A'=A$. One can find a proof in Raynaud: Anneaux locaux henséliens, IV, §3. For discrete valuation rings, one can certainly find easier references. Now let us prove the claim above $R^h$. As $\hat{R}$ is henselian (Hensel Lemma), we have $R\subseteq \hat{R}$. (1) For any algebraic separable extension $F/K$ contained in $\hat{K}$, $B_F=\hat{R}\cap F$ is a DVR with ramification index $1$ over $R$ and trivial residue extension. This is easy. (2) Let $B=B_L$ where $L$ is the separable closure of $K$ in $\hat{K}$. Let us prove that $B$ is henselian using Property (a) above. Let $f(X)\in B[X]$ as in (a). Then any simple root of $\bar{f}(X)\in k[X]$ lifts to a root $\lambda\in \hat{K}$. It is automatically a simple root, so $\lambda$ is separable over $L$, hence $\lambda\in L\cap \hat{R}=B$. (3) Let $R\to R'$ be an extension to a henselian DVR. Let's us prove it factorizes throught $R\to B$, which will show that $B$ is a henselization of $R$. It is enough to prove this factorization for $B_F$ for any finite separable exension $F/K$ contained in $\hat{K}$. As $B_F$ is an étale extension of $R$, $B_F\otimes_R R'$ is étale over $R'$, with a unique maximal ideal above the maximal ideal of $R'$ and trivial residue extension at this maximal ideal (because the quotient $B_F\otimes_R R'/(m')\simeq R'/m'$). By Property (b), this implies that $F\otimes_K K'$ has a direct factor equal to $K'$. Hence $F\subseteq K'$ (the minimal polynomial of a primitive element of $F$ has a root in $K'$) and $B_F\subseteq R'$. Remark Parts (1)(2) work for any local ring $R$ such that $\hat{R}$ is a domain and such that $R\to \hat{R}$ is injective. In particular $\hat{R}\cap K^{sep}$ is henselian. 

