Let $T$ be normal. By the spectral theorem, there is a unitary map $V : H \to L^2(\mu)$ for a suitable measure $\mu$ on a measure space $X$ such that we have
$$
T = V^\ast M_f V
$$
for some bounded function $f : X \to \Bbb{C}$, where $M_f : L^2(\mu) \to L^2 (\mu), g \mapsto f\cdot g$ is the multiplication operator which multiplies by $f$.
By the usual properties of the spectral calculus, we have $\varphi(T) = V^\ast M_{\varphi \circ f} V$ for every bounded measurable $\varphi : \Bbb{C} \to \Bbb{C}$. In particular, we have $|T|= V^\ast M_{|f|} V$.
Now, it is easy to see that there is a measurable $g : X \to \Bbb{C}$ with $|g(x)| = 1$ for all $x \in X$ and $|f(x)| \cdot g(x) = f(x)$. In fact, the function
$$
g : X \to S^1 \subset \Bbb{C}, x \mapsto \begin{cases} \frac{f(x)}{|f(x)|}, & f(x) \neq 0,\\
1, & f(x) = 0\end{cases}
$$
does the job.
Since $g$ has modulus one, the multiplication operator $M_g$ is unitary. Hence, so is $U = V^\ast M_g V$ and we have
$$
T = V^\ast M_{f} V = V^\ast M_g M_{|f|} V = V^\ast M_g VV^\ast M_{|f|} V = U\cdot |T|
$$
as desired.
EDIT: As I wrote in the comments, the unitary $U$ is not unique in general. For example for $T=0$, we have $|T|=0$, so that any unitary $U : H \to H$ will do the job.
Since $T^2=I$, you have $T^{-1}=T$. From $T=U|T|$, we get $U=T|T|^{-1}$. Then
$$
|T|^{-1}=(T^*T)^{-1/2}=[(T^*T)^{-1}]^{1/2}=[T^{-1}{T^{-1}}^*]^{1/2}=(TT^*)^{1/2}.
$$
Now the key fact is that $T(T^*T)^n=(TT^*)^nT$, and as the square root is a limit of polynomials with no constant term,
$$
T(T^*T)^{1/2}=(TT^*)^{1/2}T.
$$
Thus
$$
|T|^{-1}T=(TT^*)^{1/2}T=T(T^*T)^{1/2}=T|T|.
$$
Thus
$$
U^2=T|T|^{-1}T|T|^{-1}=TT|T||T|^{-1}=I.
$$
Best Answer
Edit: No, it does not hold in general. For some reason I always tended to use that $V$ commuted with $X^*$, which in general doesn't have to be true. A counterexample: Let $\mathcal{H}=\ell^2(\mathbb{N})$. Let $V$ be the right shift on $\mathcal{H}$, i.e. $V\delta_n=\delta_{n+1}$ where $\delta_n(m)=\left\{\begin{matrix} 1 & \mbox{if }n=m\\0 & \mbox{if }n \neq m\end{matrix}\right.$. Let $X=2+V$ (so $X\delta_n=2\delta_n+\delta_{n+1}$) then $XV=VX$. However, you can check that $X^*V \neq VX^*$. Now $X$ is invertible, so $U$ is unitary. But then $UV=VU \Leftrightarrow X^*V=VX^*$, which is false.
So the following only holds if $V$ commutes with $X^*$: Let $\mathcal{X}$ be the von Neumann algebra generated by $X$. One can show that in this case $U\in \mathcal{X}$. As clearly $V$ is an element of the commutant of $\mathcal{X}$ and $U \in \mathcal{X}$, we find that $VU=UV$. So I don't think you need the fact that $V$ is an isometry...