Details on the bound $o(d^{-100})$ in \eqref{57}: Since $y$ has the $\chi^2$ distribution with $d-1$ degrees of freedom (which is the gamma distribution with parameters $(d-1)/2$ and $2$), for each small enough $h>0$ and all $d\ge4$ we have \begin{equation} E\frac1y \,1(y<d/10)\le E\frac1y \,e^{h(d/10-y)} =e^{hd/10}\frac{(1+2 h)^{(3-d)/2}}{d-3}\le 2e^{-hd/2}=o(d^{-100}). \end{equation} Also, \begin{equation} Ez^2\,1(d/10<z^2)\le Ez^2\,e^{(z^2-d/10)/4}=O(e^{(-d/10)/4})=o(d^{-100}). \end{equation} Also, $EZ^2=1$$Ez^2=1$ and $E\frac1y=\frac1{d-3}\le1$ for $d\ge4$. So, the bound $o(d^{-100})$ in \eqref{57} follows.
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(Details on the last bound will be given at the end of this answer.)
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*}
Details on the bound $o(d^{-100})$ in \eqref{57}: Since $y$ has the $\chi^2$ distribution with $d-1$ degrees of freedom (which is the gamma distribution with parameters $(d-1)/2$ and $2$), for each small enough $h>0$ and all $d\ge4$ we have \begin{equation} E\frac1y \,1(y<d/10)\le E\frac1y \,e^{h(d/10-y)} =e^{hd/10}\frac{(1+2 h)^{(3-d)/2}}{d-3}\le 2e^{-hd/2}=o(d^{-100}). \end{equation} Also, \begin{equation} Ez^2\,1(d/10<z^2)\le Ez^2\,e^{(z^2-d/10)/4}=O(e^{(-d/10)/4})=o(d^{-100}). \end{equation} Also, $EZ^2=1$ and $E\frac1y=\frac1{d-3}\le1$ for $d\ge4$. So, the bound $o(d^{-100})$ in \eqref{57} follows.
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*}
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*} (Details on the last bound will be given at the end of this answer.)
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*}
Details on the bound $o(d^{-100})$ in \eqref{57}: Since $y$ has the $\chi^2$ distribution with $d-1$ degrees of freedom (which is the gamma distribution with parameters $(d-1)/2$ and $2$), for each small enough $h>0$ and all $d\ge4$ we have \begin{equation} E\frac1y \,1(y<d/10)\le E\frac1y \,e^{h(d/10-y)} =e^{hd/10}\frac{(1+2 h)^{(3-d)/2}}{d-3}\le 2e^{-hd/2}=o(d^{-100}). \end{equation} Also, \begin{equation} Ez^2\,1(d/10<z^2)\le Ez^2\,e^{(z^2-d/10)/4}=O(e^{(-d/10)/4})=o(d^{-100}). \end{equation} Also, $EZ^2=1$ and $E\frac1y=\frac1{d-3}\le1$ for $d\ge4$. So, the bound $o(d^{-100})$ in \eqref{57} follows.
$\newcommand{\D}{\overset D=}\newcommand{\de}{\delta}$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; see also fedja's comment.
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 $d\to\infty$.
For an illustration, below are the (linearly interpolated) plots of the values of $E\ln R$ for $d=20,\dots,100$ with $c=\frac2{d+1}$ (red) and $c=\frac2{d+2}$ (blue). These plots suggest that the threshold value of $c$ is between $\frac2{d+2}$ and $\frac2{d+1}$.
$\newcommand{\D}{\overset D=}\newcommand{\de}{\delta}$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; see also fedja's comment.
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 $d\to\infty$.
$\newcommand{\D}{\overset D=}\newcommand{\de}{\delta}$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; see also fedja's comment.
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 $d\to\infty$.
For an illustration, below are the (linearly interpolated) plots of the values of $E\ln R$ for $d=20,\dots,100$ with $c=\frac2{d+1}$ (red) and $c=\frac2{d+2}$ (blue). These plots suggest that the threshold value of $c$ is between $\frac2{d+2}$ and $\frac2{d+1}$.