$\newcommand{\D}{\overset D=}\newcommand{\de}{\delta}$Yes, the gap in condition (1) in the OP goes to $0$ as $d\to\infty$. 

To prove this, let us borrow the approach suggested by [Anthony Quas][1]; see also [fedja's comment][2].  

Let $x$ be a random vector that is a copy of $x_0$ in distribution, and let $v$ be any random vector independent of $x$. Let $Q_v$ be a random orthogonal matrix depending only on $v$ such that $Q_vv=\|v\|e_1$, where $e_1=[1,0,\dots,0]^T$. Let $y:=Q_vx$, so that $y$ is a copy of $x$ in distribution. Then 
\begin{equation*}
	\|(I-xx^T)v\|=\|Q_v^T(I-yy^T)Q_vv\|=\|(I-yy^T)e_1\|\,\|v\|=R_x\,\|v\|,
\end{equation*}
where $R_x:=\|(I-yy^T)e_1\|$ is a random variable (r.v.) independent of $v$ and equal $\|(I-xx^T)e_1\|$ in distribution. 
So, for any natural $n$ and any vector $u$, 
\begin{equation*}
	\Big\|\prod_{i=0}^n(I-x_ix_i^T)u\Big\|=\|u\|\,\prod_{i=0}^n R_{x_i},
\end{equation*}
where the $R_{x_i}$'s are independent copies of $R:=R_{x_0}$. 

So, by the strong law of large numbers, $\prod_{i=0}^\infty(I-x_ix_i^T)=0$ almost surely (a.s.) iff 
\begin{equation*}
	E\ln R<0. \tag{10}\label{10}
\end{equation*}

Letting $z=[z_1,\dots,z_d]^T:=x/\sqrt c$, we get independent standard normal $z_i$'s, and 
\begin{equation*}
	R^2=(1-c z^2)^2+c^2 z^2 y, \tag{20}\label{20}
\end{equation*}
where $z:=z_1$ and $y:=z_2^2+\cdots+z_d^2$, so that the r.v. $y$ is independent of the standard normal r.v. $z$ and has the $\chi^2$ distribution with $d-1$ degrees of freedom. 

Note that $ER^2\le1$ iff $(d+2)c\le2$. Therefore and because $\ln(R^2)<R^2-1$ if $R\ne1$, we see that the condition $(d+2)c\le2$ or, equivalently, 
\begin{equation}
	c\le\frac2{d+2} \tag{25}\label{25}
\end{equation}
is sufficient for \eqref{10}. (Note that the strict inequality $(d+2)c<2$ is equivalent to condition (1) in the OP.) 

Note next that $R^2\ge c^2 z^2 y$ and hence 
\begin{equation*}
2	E\ln R\ge E\ln(c^2 z^2 y)=\ln(c^2)+\psi\Big(\frac{d-1}{2}\Big)-\gamma
=\ln((K-o(1))c^2d),
\end{equation*}
where $K:=e^{-\gamma}/2$, $\gamma=0.577\ldots$ is Euler's gamma, and $\psi:=\Gamma'/\Gamma$. So, for \eqref{10} it is necessary that 
\begin{equation*}
	c\le\frac C{\sqrt d}, \tag{30}\label{30}
\end{equation*}
where $C$ is a universal positive real constant; here and in what follows, $d$ is is any large enough natural number. 

Minimizing in $c$, from \eqref{20} we get 
\begin{equation*}
\text{$R^2\ge R_*^2:=\frac{y}{y+z^2}$, and $E\ln(R_*^2)>-\infty$.	} \tag{40}\label{40}
\end{equation*}

To obtain a contradiction with \eqref{10}, suppose that $c^2 d\not\to0$ (as $d\to\infty$) -- cf. \eqref{30}. Then without loss of generality $c^2 d\to\de$ for some real $\de>0$, whence, by \eqref{20} and the strong law of large numbers for $y$, we have $R^2\to1+\de\, z^2$ a.s. So, by \eqref{40} and the Fatou lemma, $\liminf_{d\to\infty}(2	E\ln R)\ge E\ln(1+\de\, z^2)>0$, which indeed contradicts \eqref{10}. So, 
\begin{equation*}
	c^2 d\to0. \tag{50}\label{50}
\end{equation*}

We have 
\begin{equation}
	\text{$\ln t\ge h(t):=t-1-k(t-1)^2$ }\\ \text{for some $k\in(0,\infty)$ and all $t\in[1/2,\infty)$.} \tag{55}\label{55}
\end{equation}
 
Next,  
\begin{equation*}
\begin{aligned}
	Eh(R^2) 
	&=c [-2 + c (2 + d - 12 k) + 12 c^2 (4 + d) k \\ 
	&\qquad\qquad- 3 c^3 (24 + 10 d + d^2) k] \\ 
	&= c (-2+o(1) + (1+o(1))c d) \\ 
		&=cd\,\Big(-\frac{2+o(1)}d  + (1+o(1)) c\Big),  
\end{aligned}
\tag{56}\label{56}
\end{equation*}
in view of \eqref{50}. 

Next, using \eqref{40} again, we get 
\begin{equation*}
\begin{aligned}
&	E\ln(R^2) \,1\Big(\frac y{y+z^2}<\frac12\Big) \\ 
&\ge E\ln\frac y{y+z^2} \,1\Big(\frac y{y+z^2}<\frac12\Big)  \\ 
&\ge -E\frac{z^2}y \,1\Big(\frac y{y+z^2}<\frac12\Big)  \\ 
&=-E\frac{z^2}y \,1(y<z^2)  \\ 
&\ge-E\frac{z^2}y \,1(y<d/10)-E\frac{z^2}y \,1(d/10<z^2)  \\ 
&=-Ez^2\,E\frac1y \,1(y<d/10)-E\frac1y \,Ez^2\,1(d/10<z^2) \\
&=-o(d^{-100}).  
\end{aligned}
\tag{57}\label{57}
\end{equation*}
Similarly, 
\begin{equation*}
\begin{aligned}
&	Eh(R) \,1\Big(\frac y{y+z^2}<\frac12\Big) \\ 
&\le E(R^2-1) \,1(y<z^2)  \\ 
&\le2c^2 Ez^4\,1(y<z^2) \le Ez^4\,1(y<z^2)=o(d^{-100}).  
\end{aligned}
\tag{58}\label{58}
\end{equation*}

Collecting \eqref{10}, \eqref{55}, \eqref{40}, \eqref{57}, \eqref{58}, and \eqref{56}, we get 
\begin{equation*}
\begin{aligned}
&	0>E\ln(R^2) \\ 
&=E\ln(R^2)\,1\Big(\frac y{y+z^2}\ge\frac12\Big)+E\ln(R^2)\,1\Big(\frac y{y+z^2}<\frac12\Big) \\  
&\ge Eh(R^2)\,1\Big(\frac y{y+z^2}\ge\frac12\Big)-o(d^{-100}) \\  
&\ge Eh(R^2)-o(d^{-100})-o(d^{-100}) \\  
&=cd\,\Big(-\frac{2+o(1)}d  + (1+o(1)) c\Big)-o(d^{-100}).  
\end{aligned}
\end{equation*}

So, 
\begin{equation*}
	c\le\frac{2+o(1)}d \tag{60}\label{60}
\end{equation*}
is necessary for \eqref{10}. Since \eqref{25} was shown to be sufficient for \eqref{10}, we conclude that \eqref{60} is necessary and sufficient for \eqref{10}. 

But \eqref{60} is equivalent to $c(d+2)\le2+o(1)$, which can be rewritten as  $\operatorname{Tr}\Sigma +2\|\Sigma\|\le2+o(1)$. So, indeed the gap in 
condition (1) in the OP goes to $0$ as $d\to\infty$. 

[1]: https://mathoverflow.net/a/430790/36721  
[2]: https://mathoverflow.net/questions/430781/range-of-a-such-that-w-leftarrow-w-a-x-langle-w-x-rangle-converges-almos/430790#comment1143873_430790