No, a general set of 10 points in $\mathbb P^3$ do not form the singular locus of a symmetroid. This goes back to the papers of Cayley on quartic surfaces. If I remember correctly, to construct a symmetroid set, one can fix 7 of the points in $\mathbb P^3$ freely, then the 8th can be chosen in a 2-dimensional family $S$ (a "Cayley surface"), and the ninth along a curve on $S$ (and then the 10th point is uniquely determined for it to be a symmetroid). This is explained in Jessop's book 'Quartic surfaces with singular points'.
One can see that the nodes are special position as follows. Given a quartic surface $X\subset\mathbb P^3$ with nodes at $p_0,\ldots,p_9$ one can project from one of the nodes, say $p_0$ to get a double cover of the plane branched along a degree 6 curve $C$. If $X$ is a symmetroid, one can check that this curve is reducible and is generically a union of two cubic curves. It was showed by Cayley (and more recently in this paper) that also the converse holds - a quartic is a symmetroid if and only if the degree 6 curve splits into two cubics. Note that the images of the points $p_1,\ldots,p_9$ are the intersection of these cubics. So these 9 points are special in the sense that there is a pencil of cubics passing through them.