No. The construction of Mitchell which is described in the paper by Laver and Shelah as indicated in the comments by Golshani can be modified to produce a counterexample to your question. I had to answer before Mohammad does so that this chain grows even longer.
Claim: Assuming CH and $\diamondsuit_{E^{\omega_2}_{\omega_1}}$, there is a sequenence $\langle T_n\mid n<\omega\rangle$ so that $\prod_{i<n} T_i$ is a countably closed $\omega_2$-Suslin tree for all $n<\omega$, but the (full support) product $\prod_{n<\omega} T_n$ collapses $\omega_2$.
This immediately gives a counterexample assuming CH + $\diamondsuit_{E^{\omega_2}_{\omega_1}}$, namely the iteration along the $T_i$. Note that this is the same as the full support product of the $T_i$ as all $T_i$ are countably closed. (This is not the case in general as the full support product of $\omega$-copies of Cohen forcing is very different from the full support iteration)
From this, we get a ZFC (+CH) counterexample as we can first force with $\mathrm{Add}(\omega_2, 1)$ to get the relevant $\diamondsuit$-principle.
So let me sketch the construction of $\mathcal T:=\prod_{n<\omega}T_n$. We construct $\mathcal T$ inductievly level-by level.
At the same time, we construct $c^\alpha_\xi\in\mathcal T_\xi$ for $\alpha<\omega_2$, $\xi\in E^{\omega_2}_{\omega_1}\setminus \alpha$ so that for all $\alpha<\omega_2$,
$\mathcal A^\alpha=\{c^\alpha_\xi\mid \xi\in E^{\omega_2}_{\omega_1}\setminus\alpha\}$ ends up as a maximal antichain of $\mathcal T$. At the same time, we make sure that we keep $\mathcal T$ a normal tree and keep two inductive assumptions alive which we will discuss as we go.
First, let me describe the mechanism by which $\mathcal T$ collapses $\omega_2$, as this motivates the construction:
Every $d\in \mathcal T$ defines a maximal continuous increasing sequence $\mathrm{seq}(d)\colon\beta\rightarrow\omega_2$ via $\mathrm{seq}(d)(0)=0$ and $\mathrm{seq}(d)(\gamma+1)=$ $\mathrm{seq}(d)(\gamma)+\xi$ where $\xi$ is unique with $c^{\mathrm{seq}(d)(\gamma)}_\xi\leq d$. Such a $\xi$ may not exist in which case $\beta=\gamma+1$. We will arrange that $\mathrm{dom}(\mathrm{seq}(d))$ is countable for all $d\in\mathcal T$ and if we force with $\mathcal T$ then the $\mathrm{seq}(d)$ for $d\in G$ will union up to a cofinal function from $\omega_1$ to $\omega_2$, so $\omega_2$ is collapsed.
At successor steps, we give every node of any $T_n$ exactly two immediate successors and at limit steps of countable cofinality, every cofinal branch is extended.
At a limit step $\mu$ of cofinality $\omega_1$, we are handed a maximal antichain $A\subseteq \prod_{i<n} (T_i)_{<\mu}$ for some $n<\omega$ by the $\diamondsuit_{E^{\omega_2}_{\omega_1}}$-sequence. We will make sure that we put $c=\langle c_i\mid i<\omega\rangle$ into $\mathcal T_\mu$ only if
$\langle c_i\mid i<n\rangle$ is above a point of $A$ and
$\mathrm{dom}(\mathrm{seq}(c))$ is countable.
We have to show that there are enough of such $c$ so that we can keep our inductive assumptions going. Given any $x=\langle x_i\mid i<\omega\rangle\in\mathcal T_{<\mu}$, we may assume that $\langle x_i\mid i<n\rangle\in A$ as we can strengthen $x$ otherwise. By induction, we have that $\beta=\mathrm{dom}(\mathrm{seq})(x)<\omega_1$, so $\beta=\gamma+1$. The crucial step now is that there is a cofinal branch $b\subseteq\mathcal T_{<\mu}$ containing $x$ so that for all $y\in b$ above $x$ we have $\mathrm{dom}(\mathrm{seq}(y))=\beta$ as well: Let $f\colon\omega_1\rightarrow E^\mu_\omega$ enumerate a club in $\mu$ with $x\in \mathcal T_{<f(0)}$. $x$ is not above any $c^{\mathrm{seq}(x)(\gamma)}_\xi$ and we can construct increasing $x_i\in\mathcal T_{f(i)}$ for which this still holds true. This does not run into trouble at limit steps $i$ as we have $\mathrm{cof}(f(i))=\omega$ so there is no point of $\mathcal A^{\mathrm{seq}(x)(\gamma)}$ on level $f(i)$ to worry about. To get through the successor steps we use our second inductive assumption:
For any $\delta_0<\delta_1<\mu$, $\alpha_0, \alpha_1<\mu$, $a=\langle a_i\mid i<\omega\rangle\in \mathcal T_{\delta_0}$ so that $a$ is not above any $c^{\alpha_j}_\xi$ with $\xi\in E^\mu_{\omega_1}\setminus\alpha_j$, $j<2$, and any $\langle b_i\mid i<n\rangle\in \prod_{i<n} (T_i)_{\delta_1}$ above $\langle a_i\mid i<n\rangle$, there is a node $b=\langle b_i\in i<\omega\rangle\in\mathcal T_{\delta_1}$ above $a$ which is not above any $c^{\alpha_j}_\xi$ with $\xi\in E^\mu_{\omega_1}\setminus\alpha_j$, $j<2$.
The extra flexibility will become important shortly.
It remains to define $c^\alpha_\mu$ for $\alpha\leq \mu$. To make sure that $\mathcal A^\alpha$ will be maximal, some bookkeeping hands us some $x\in \mathcal T_{<\mu}$ not above any $c^\alpha_\xi$ with $\xi\in E^\mu_{\omega_1}\setminus\alpha$ and we make sure that $c^\alpha_\mu$ is above $x$. Also, $c^\alpha_\mu$ has to satisfy 1. + 2. as well. It is at this point that we use the second inductive assumption to its full extend. First, we may assume wlog that $x\upharpoonright n$ is above a point in $A$. We can then run the same argument as before, only that now we have to worry to not get above neither any $c^\alpha_\xi$ nor any $c^{\max\mathrm{ran}(\mathrm{seq}(x))}_\xi$. We are saved once again by the inductive assumption.
