Is the statement that $U(x)$ is quadratic for simple harmonic motion equally strong as the statement that $F(x)$ is linear?

Both of your statements are true for SHM and can be seen as true for approximations as well. Each statement is essentially saying the same thing, because $F=-\text dU/\text dx$, and we can choose $U(0)=0$. Whether or not you use approximations to make these statements applicable to your system doesn't change that.

In other words, if you have approximated $U\propto x^2$, then you are also approximating $F\propto -x$. Both of your statements say the same thing. Therefore, they are equally "strong".

Let's consider your first statement. If the potential energy of a particle under oscillatory motion is directly proportional to the second power of displacement from the mean position, then we can write the expression for potential at any position $x$ as $$U(x-x_0)=A(x-x_0)^2$$ where $A$ is our proportionality constant and $x_0$ is the mean position, which makes $x-x_0$ the displacement from the mean position. After expanding $$U(x-x_0)=Ax_0^2-2Ax_0x+Ax^2.$$ Now we know that $F=-{dU(x-x_0) \over dx}$. So, for our potential energy, it becomes- $$F=-(-2Ax_0+2Ax)=2Ax_0-2Ax=-2A(x-x_0)$$ which mirrors your second statement. So they're equivalent for a single dimension. And we get those two statement equally strong in Euclidean space of arbitrary dimension. Verify it. But let's not stop here.
Consider your first statement in a non-Euclidean space. In this case, the expression for our potential is $U=A ({\sqrt {g_{\mu \nu}dx^\mu dx^\nu}})^2=Ag_{\mu \nu}dx^\mu dx^\nu$. According to your second statement, we expect $F^\mu=-Bdx^\mu$.
According to the definition of force, $$F^\mu=-g^{\mu \nu}\nabla_\nu U=-g^{\mu \nu}\partial_\nu(Ag_{\sigma \rho}dx^\sigma dx^\rho)=-2Adx^\mu-Ag^{\mu \nu}g_{\sigma \rho, \nu}dx^\rho dx^\sigma.$$So, if we want this last expression to meet our expectation, we must impose one condition, that is the value of $g_{\sigma \rho,\nu}$ must be zero.

So, it seems that for the later case your second statement impose more restriction than your first one. So your first statement is more general, but the second one is stronger.
Plea: Can someone please verify my claims and calculation for the non-Euclidean part?