No, the completion of a strictly convex normed space can fail to be strictly convex.
To put it differently, there are non strictly convex Banach spaces with a dense strictly convex subspace.
Here is a possible construction. To make things easier, it is tempting to start with a space where there is a good control on the non strictly convex part. For example, let $E,F$ be you favorite separable infinite dimensional strictly convex Banach spaces (for example $\ell_2$), and consider $E\oplus_{\ell_1} F$, that is the direct sum of $E$ and $F$ for the norm $\|(v_1,v_2)\| = \|v_1\|_E + \|v_2\|_F$. In that way, for $v=(v_1,v_2)$ and $w=(w_1,w_2)$, $\|v+w\| = \|v\|+\|w\|$ if and only if for each $i \in \{1,2\}$, $v_i$ and $w_i$ are positively proportional. But if $v$ and $w$ are not proportional, this implies that there is a nonzero linear combination of $v$ and $w$ that is in $0\oplus F$ or $E\oplus 0$. In other words, any subspace $V \subset E\oplus_{\ell_1} F$ which interesects trivially $E \oplus 0$ and $0 \oplus F$ is strictly convex.
But it is a small exercise that, given a finite number of infinite codimensional subspaces $E_i$ of a separable Banach space, there is a dense subspace which does intersects them all trivially.
(solution: if $(x_n)$ is a dense sequence, construct by induction a sequence $(y_n)$ such that $\|x_n-y_n\| \leq \frac 1 n$ and $\operatorname{span}(y_1,\dots,y_n) \cap E_i = \{0\}$ for all $i$; the space spanned does the job).
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