The answer is 'not always', in particular, not when $(n,r)=(2,3)$.
The image is obviously is a smooth manifold when $n=0$, for then the image in $\mathbb{R}^{(n+1)r}=\mathbb{R}^r$ is the sphere $\Sigma$ of radius $\sqrt{r}$ centered on the origin in $\mathbb{R}^r$.
The image (perhaps less obviously) is a smooth manifold when $n=1$, for then the image in $\mathbb{R}^{(n+1)r}=\mathbb{R}^{2r}$ is $\Sigma\times\mathbb{R}^r$ in $\mathbb{R}^{2r}$.
The image is a smooth manifold when $r=1$ (and $n$ is arbitrary), since it is the union of the two curves $\mathcal{V}$ and $-\mathcal{V}$.
The image $\mathcal{VU}$ is a smooth, connected $3$-manifold in $\mathbb{R}^{2(n+1)}$ diffeomorphic to $S^1\times \mathbb{R}^2$ when $r=2$: In fact, for $n>1$ it can be written as a smooth graph over the image when $n=1$, as is easy to check.
However, when $n=2$ and $r=3$ the image $\mathcal{VU}$ is not a smooth submanifold of $\mathbb{R}^{(n+1)r}\simeq\mathbb{R}^{9}$, and I suspect that, except for the simple cases listed above, it's never a smooth submanifold again. I haven't checked for all of them, but in the $(n,r)=(2,3)$ case, one can show directly that the image is not a smooth submanifold in $\mathbb{R}^{9}$.
Here is an outline of the argument. For clarity, I'm going to use the notation $\mathbb{R}^p_q$ to denote the vector space of $p$-by-$q$ matrices with entries from $\mathbb{R}$, and, when $p$ or $q$ is equal to $1$, I'll omit it. Thus, $\mathbb{R}^r$ is the vector space consisting of a columns of height $r$ of real numbers, and $\mathbb{R}_r$ is the space of rows of length $r$. For $m\in\mathbb{R}^p_q$, I will let $m^*\in \mathbb{R}_p^q$ denote its transpose.
Thus, the orthogonal group $\mathrm{O}(r)$ is the Lie group of matrices $a\in\mathbb{R}^r_r$ that satisfy $a^*a = I_r$. For explicitness, I will let $u:\mathrm{O}(r)\hookrightarrow \mathbb{R}^r_r$ denote the inclusion mapping. There is a unique $\mathbb{R}^r_r$-valued $1$-form $\theta$ on $\mathrm{O}(r)$ that satisfies $\theta^* = -\theta$ and $\mathrm{d}u = \theta\,u$, which will come in handy below.
It will be useful to let $S_r\subset \mathrm{O}(r)$ denote the subgroup of permutation matrices, i.e., elements of $\mathrm{O}(r)$ that have a single $1$ in each row and column while all the other entries are $0$. The order of $S_r$ is $r!$.
Let $V:\mathbb{R}^r\to\mathbb{R}^{n+1}_r$ be the Vandermonde mapping as defined in the question. Note that $V(px) = V(x)p^*$ for all $x\in \mathbb{R}^r$ and $p\in S_r$.
We are interested in the image of the smooth mapping $F:\mathbb{R}^r\times\mathrm{O}(r)\to\mathbb{R}^{n+1}_r$ defined by $F(x,a) = V(x)a$. It is worth pointing out that
$$
F(px,pa) = V(x)p^*pa = V(x)a = F(x,a),
$$
so that $F$ induces a well-defined mapping $\bar F:X_r\to \mathbb{R}^{n+1}_r$ on the (smooth) quotient manifold $X_r = S_r\backslash\bigl(\mathbb{R}^r\times\mathrm{O}(r)\bigr)$, which has dimension $\tfrac12(r^2{+}r)$ and is connected when $r\ge2$. It's also worth pointing out that
$$
Q(x) = F(x,a)F(x,a)^* = V(x)a a^* V(x)^*
= V(x)V(x)^* = \bigl(q_{ij}(x)\bigr),
$$
where $q_{ji}(x) = q_{ij}(x) = p_{i+j}(x)$ where $p_k(x) = {x_1}^{k} + \dots + {x_r}^{k}$. Of course, $p_0(x)=r$, and, as is well known, the quantities $p_1(x),\ldots, p_r(x)$ determine the $r$ entries of $x\in\mathbb{R}^r$ up to permutation. Thus, when $2n\ge r$, if $F(x,a) = F(y,b)$, then there exists a $p\in S_r$ such that $y = px$, implying that $F(x,a)=F(y,b)=F(px,b) = F(x,p^*b)$, so that $V(x)(a-p^*b)=0$. In particular, if $n\ge r{-}1$ and the $r$ entries of $x$ are distinct, then $p^*b = a$, implying that $(y,b) = (px,pa)$. Consequently, the induced map $\bar F:X_r\to\mathbb{R}^{n+1}_r$ is injective on the locus corresponding to the dense open set where $x$ has $r$ distinct entries. Thus, when $n\ge r{-}1$, the image of $\bar F$ (which is the image of $F$), is a connected algebraic variety of dimension $\tfrac12r(r{+}1)$. Also, note that, when $n\ge 1$, the fact that $Q(x)$ contains the entry $q_{11} = x^*x = |x|^2$ implies that $F$ is a proper mapping; in particular, its image is closed.
Regarding $F$ as a matrix-valued function, one can compute its exterior derivative as
$$
\mathrm{d}F = \mathrm{d}(Vu) = \mathrm{d}V\,u+ V\,\mathrm{d}u
= (\mathrm{d}V+V\theta)u.
$$
Since $u$ is invertible, the rank of $\mathrm{d}F$ at a point $(x,a)$ is the same as the number of linearly independent entries of $\mathrm{d}V{+}V\theta$ at $(x,a)$. When $n\ge 1$, this rank is at least $2r{-}1$ (just look at the top two rows), and, when $n\ge r{-}1$ and the $r$ entries of $x$ are distinct, this rank is the maximum $\tfrac12r(r{+}1)$.
Now, consider the particular case $(n,r)=(2,3)$. By explicit calculation, one finds that the rank of the differential $\mathrm{d}V{+}V\theta$ is equal to its maximum value of $6$ except at points $x$ where $x_1=x_2=x_3$, where it drops to $5$. Thus, the image of $F$ is a smooth $6$-manifold except possibly at the points that are the image of $(x,a)=\bigl((t\ t\ t)^*,a\bigr)$.
Since $r=3$, $p_4(x)$ can be expressed as a polynomial in $p_1(x)$, $p_2(x)$, and $p_3(x)$:
$$
p_4(x) = \tfrac43 p_3(x)p_1(x) + \tfrac12\bigl(p_2(x)\bigr)^2
-p_2(x)\bigl(p_1(x)\bigr)^2 + \tfrac16\bigl(p_1(x)\bigr)^4.
$$
Thus, letting $f_0(x,a)$, $f_1(x,a)$, and $f_2(x,a)$ denote the three rows of $F(x,a)$, one finds that these rows must satisfy three relations: $f_0{f_0}^*-3 = 0$, $f_1{f_1}^*-f_0{f_2}^*=0$, and
$$
f_2{f_2}^*-\tfrac43\,(f_1{f_2}^*)(f_0{f_1}^*)-\tfrac12\,(f_1{f_1}^*)^2+(f_1{f_1}^*)(f_0{f_1}^*)^2-\tfrac16(f_0{f_1}^*)^4
= 0.
$$
These define three independent polynomials of degrees $2$, $2$, and $8$ on $\mathbb{R}^3_3$ whose common zero locus $Z$ (which has dimension $6$) contains the image of $F$. It is now not difficult to verify that $Z$ is not smooth at a point of the form $F\bigl((t\ t\ t)^*,a\bigr)$. In fact, there is no smooth $6$-dimensional submanifold contained in $Z$ that passes through such a point. Hence, the image of $F$ cannot be a smooth submanifold of dimension $6$ in a neighborhood of such a point (though it is smooth and of dimension $6$ away from such points).
