Your answer is correct, if "unknotted" is equivalent to: being homeomorphic to a union of standard sphere complements within disjoint convex balls. Philosophically, you could imagine giving a proof by saying that such a space is a connected sum of sphere complements, and understand enough about them to prove it inductively. Here is an argument that is a little more homotopy-theoretic, which calculates the answer inside a closed ball of large radius and then gets the answer for a sphere by gluing on a disc.

You need a basic calculation to get started. For $k \leq n-2$, the complement $C_k$ of the standard k-sphere (embedded in the first $(k+1)$ coordinates) in the standard n-ball of radius 2 has an explicit determination of its homotopy type. It has a deformation retraction down to the union of the (n-1)-sphere of radius 2, together with the $(n-k-1)$-dimensional ball of radius 2 in the last $(n-k-1)$ coordinates. (This is a "draw a straight line from the nearest point on the $k$-sphere" retraction, I believe.) This space formed by attaching an $(n-k-1)$-dimensional cell along a map $S^{n-k-2} \to S^{n-1}$; this is necessarily nullhomotopic, so the space is homotopy equivalent to $S^{n-1} \vee S^{n-k-2}$, as the homology suggests. Under this, the inclusion from the outside boundary is homotopic to the inclusion of the first wedge factor.

The complement $Y$ of $j$ nonintersecting convex balls in a closed ball of radius $R$ is homotopy equivalent to the complement of $j$ points in $R^n$, and this makes it homotopy equivalent to $\bigvee_{i=1}^{n} S^{n-1}$. We can use a degree calculation in homology to be more explicit: the inclusion of each boundary component of one ball is homotopic to the inclusion of a wedge factor, while the inclusion of the outside boundary is homotopic to the "pinch" map $S^{n-1} \to \bigvee S^{n-1}$.

The complement $Z$ of $k_1$ 1-spheres, $k_2$ 2-spheres, etc, in a closed ball of radius $R$ is the pushout:
$$
Y \leftarrow \coprod S^{n-1} \rightarrow \coprod_i \left(\coprod^{k_i} C_{k_i}\right).
$$
Up to homotopy equivalence, the previous identifications say that we have the pushout of
$$
\bigvee^{\sum k_i} S^{n-1} \leftarrow \coprod^{\sum k_i} S^{n-1} \rightarrow \coprod \left(\coprod^{k_i} S^{n-1} \vee S^{n-1-k_i}\right),
$$
with the map on the left collapsing basepoints and the map on the right being the coproduct of wedge inclusions. The result is that we identify basepoints in all the coproduct factors, and so the resulting space is homotopy equivalent to
$$
\bigvee \left(\bigvee^{k_i} S^{n-1} \vee S^{n-1-k_i}\right) \cong \bigvee^{\sum k_i} S^{n-1} \vee \bigvee \left(\bigvee^{k_i} S^{n-1-k_i}\right).
$$
Moreover, the inclusion of the outside boundary sphere is the fold map from $S^{n-1}$ to the first wedge of $(n-1)$-spheres.

Now finally, the complement $X$ of $k_1$ 1-spheres, etc, in the $n$-sphere, is the pushout
$$
Y \leftarrow S^{n-1} \rightarrow D^n
$$
where the left-hand map is the inclusion of the outside boundary sphere of radius $R$. Up to homotopy equivalence, this is the pushout of
$$
\bigvee^{\sum k_i} S^{n-1} \vee \bigvee \left(\bigvee^{k_i} S^{n-1-k_i}\right) \leftarrow S^{n-1} \rightarrow D^{n-1}.
$$
The left-hand map is the pinch map into the first wedge factor. Up to homotopy equivalence, gluing in this cell eliminates one sphere factor from the wedge; the resulting space is the one you were hoping for.