The answer is yes. In fact, we don't need the spectral theorem to prove it.
Suppose that $M$ is a real invertible matrix. Then $M^TM$ is positive definite and has the unique positive semidefinite square root $P = \sqrt{M^TM}$. Now, note that $P$ has the property $\|Px\| = \|Mx\|$ for all vectors $x$.
If $M$ is invertible, then $P$ is invertible as well, and we have $M = (MP^{-1})P$. We note that $MP^{-1}$ is orthogonal since for all $y = Px$, we have
$$
\|MP^{-1}y\| = \|y\|
$$
Thus, we have a polar decomposition with $O =( MP^{-1})$ .
We note that this decomposition is unique. In particular, suppose $M = OP$ for orthogonal $O$ and positive $P$. then
$$
M^TM = PO^TOP = P^2
$$
And so, by the uniqueness of positive definite square roots, $P$ is uniquely determined. Then we can rearrange $M = OP$ to find $O = MP^{-1}$ is also unique determined.
Things get a bit trickier when $M$ is not invertible, but we can still guarantee a (non-unique) polar decomposition.
If $A=USV^\ast$ is a singular value decomposition of a non-normal traceless $2\times2$ matrix $A$, then $V^\ast U$ must possess a zero diagonal.
Write $-\det(A)$ in polar form as $de^{i\theta}$. By dividing $A$ by $e^{i\theta/2}$ and by a change of orthonormal basis, we may assume that $-\det(A)=d\ge0$ and $V=I$. We want to show that $U$ has a zero diagonal.
Since $A$ has a zero trace, $A^2=dI$. Therefore $USUS=dI$.
If $A$ is singular, then $SUS=0$. Since $A$ is not normal, $S=\operatorname{diag}(s,0)$ for some $s>0$. The equality $SUS=0$ thus implies that $u_{11}=0$. As $U$ is unitary, $u_{22}$ must also be zero. Hence $U$ has a zero diagonal.
If $A$ is nonsingular, then $d>0$. From $USUS=dI$, we get $(USU^\ast)U^2 = \left(dS^{-1}\right)(I)$. By the uniqueness of polar decompositions of nonsingular matrices, we have $U^2=I$. As $U\ne\pm I$ (otherwise $A=\pm S$ is normal), the spectrum of $U$ must be equal to $\{1,-1\}$. Hence the trace of $U$ is zero. If the diagonal of $U$ is nonzero, since $A=US$ also has zero trace, $S$ must be a scalar matrix and $A=US$ is normal, which is a contradiction. Therefore $U$ has a zero diagonal.
Best Answer
The answer is yes: it is always possible.
$U$ will satisfy $A = UP$ if and only if it satisfies $$ U(Px) = Ax \quad \forall x\in \Bbb C^n $$ that is, it sufficient to determine how $U$ acts on the image of $P$, which in your case is a strict subset of $\Bbb R^n$.
Let $\{x_1,\dots,x_r\}$ be an orthonormal basis for the image of $P$. Extend this to an orthonormal basis $\{x_1,\dots,x_n\}$ of $\Bbb R^n$. Let $V$ be a unitary transformation that, with respect to this basis, has the form $$ V = \pmatrix{I_{r \times r} &0\\0 & V'} $$ where $I$ denotes the identity matrix and $V'$ is an arbitrary unitary matrix. Then for any $U_1$ such that $A = U_1P$, $U_2 = VU_1$ will also be unitary and satisfy $A = U_2P$. As long as $V \neq I$, we will have $U_1 \neq U_2$.