I understand the question as follows: can we determine a nondegenerate quadric surface from its traces in the three coordinate planes?
In general, we cannot since all three traces may be empty - for example, take the sphere with radius $1$ centered at $(2,2,2)$. For a less trivial example, consider two paraboloids: circular $z=x^2+y^2$ and elliptic $z=x^2+y^2+xy$. They are different surfaces but we cannot tell them apart by their traces in coordinate planes. There is a similar example with one-sheeted hyperboloids: just replace $z$ with $z^2-1$ on the left.
There is a pattern in the above counterexamples: at least one of the traces was degenerate. Let's assume that all three traces are nondegenerate quadrics - that is, circles/ellipses, parabolas or hyperbolas. Can we recover the surface in this case? The answer is yes and I think I can prove it.
Case 1: the surface does not pass through the origin (which we see from its coordinate traces). Then its equation (which we don't know yet) can be brought into the form $p(x,y,z)=1$ where $p$ has no constant term. We can write the equations of traces as $f(x,y)=1$, $g(x,z)=1$ and $h(y,z)=1$. Now, $p-f$ vanishes when $z=0$, which means it only involves the terms divisible by $z$: namely, $xz$, $z^2$, and $yz$. Of these, the first two are found in $g$ and the last one in $h$. Notice that we are not adding $f,g,h$ together - we merge them by taking the union of all monomials they contain. Something very similar happens in calculus when students are asked to recover a function from its partial derivatives.
Case 2: the surface passes through the origin. Its equation is now $p(x,y,z)=0$ where $p$ still does not have constant term. The trace equations are now $f=0$, $g=0$, and $h=0$ which we can normalize so that at the origin $f_x=g_x$, $f_y=h_y$, and $g_z=h_z$. At most one of these equations may be $0=0$; thus we have enough information to determine the ratios of coefficients with which $f,g,h$ appear in $p$. (This is where nondegeneracy of traces is needed.) Now we merge $f,g,h$ into $p$ as in Case 1.