Yes, if the convex body is "sufficiently round". If it is not, the resulting "closeness" to the boundary of a convex set is in absolute terms rather than relative. I don't know whether it can be improved, but Bill Johnson's remarks suggests that it can't.

Let $X=\{x_i\}$, $i=1,\dots,k$, be the set in question and $K$ the convex body whose boundary contains $X$. For normalization assume that $diam(K)=1$. For each $i$ let $y_i$ be a point in $\mathbb R^d$ such that the vector $y_i-x_i$ equals the unit inner normal to a hyperplane supporting $K$ at $x_i$. Then
$$
 \langle y_i-x_i, x_j-x_i \rangle \ge 0
$$
for all $j$ (the angle brackets denote the scalar product). Apply the Johnson-Lindenstrauss lemma to the set $X\cup\{y_i\}$and let $x_i'$ and $y_i'$ be the images of $x_i$ and $y_i$ in $\mathbb R^{d'}$ such that all distances are preserved up to a relative (and hence absolute) error $\varepsilon$. Recall that the scalar product can be recovered from the distances by the formula
$$
 \langle y_i-x_i, x_j-x_i \rangle  = \frac12(|x_iy_i|^2+|x_ix_j|^2-|y_ix_j|^2 .
$$
Since the distances are almost preserved and the product is nonnegative, for the images we get
$$
\langle y_i'-x_i', x_j'-x_i' \rangle \ge -3\varepsilon
$$
This means that all $x_j'$ belong to the half-space $H_i$ defined by
$$
 H_i = \{ x'\in\mathbb R^{d'} : \langle x'-x_i', y_i'-x_i' \rangle \ge -3\varepsilon \}
$$
Since $|y_i'-x_i'|=1\pm\varepsilon $, the point $x_i'$ lies within distance $3\varepsilon/(1-\varepsilon)\le 4\varepsilon$ from the boundary of $H_i$ (assuming $\varepsilon\le 0.1$). Therefore the projection of $X$ is contained in the convex set $K':=\bigcap_i H_i$ and lies within distance $4\varepsilon$ from its boundary.