$\DeclareMathOperator{\Gr}{Gr}$ It is convenient to identify $G(r,n-r)$ with the space $\newcommand{\bR}{\mathbb{R}}$ $\Gr_r(\bR^n)$ of $r$-dimensional subspaces of $\bR^n$ via the map
$$
\Gr_r(\bR^n)\ni L\mapsto P_L \in G(r,n-r),
$$
where $P_L$ is the orthogonal projection on $L$. The operator $I_{r,n-r}$ corresponds to the subspace $L_0=\bR^r\oplus 0\subset \bR^n$.

Denote by $\newcommand{\eO}{\mathscr{O}}$ $\eO$ the open subset of $\Gr_r(\bR^n)$ consisting of subspaces that intersect $L_0^\perp=0\oplus \bR^{n-r}$ transversally. $\DeclareMathOperator{\Hom}{Hom}$ The open set $\eO$ can be identified with $\Hom(\bR^r,\bR^{n-r})$ via the graph map
$$
\Hom(\bR^r, \bR^{n-r})\ni S\mapsto \Gamma_S\in \eO,
$$
$$
\Gamma_S=\big\{(x,Sx)\in\bR^r\times \bR^{n-r}:\;\;x\in\bR^r\big\}.
$$
We think of matrices $S\in\Hom(\bR^r,\bR^{n-r})$ as defining coordinates on $\eO$

Using the decomposition $\bR^n=\bR^r\times\bR^{n-r}$ we can describe an $n\times n$ matrix $X$in blocks
$$
X=\left[
\begin{array}{cc}
A & B\\
C & D
\end{array}
\right],
$$
where $A$ is $r\times r$, $D$ is $(n-r)\times (n-r)$ etc.

For $S\in\Hom(\bR^r,\bR^{n-r})$ (or equivalently a $(n-r)\times r$ matrix) the orthogonal projection onto $\Gamma_S$ has the block decomposition (see Eq. (1.2.5) p.17 of this book)
$$
P_{\Gamma_S}=\left[
\begin{array}{cc}
(1+S^*S)^{-1} & (1+S^*S)^{-1}S^*\\
S (1+S^*S)^{-1} & S (1+S^*S)^{-1} S^*
\end{array}
\right].
$$
For simplicity we set $W(S):=(1+S^*S)^{-1}\in\Hom(\bR^r,\bR^r)$. The plane $L_0=\bR^r\oplus 0$ has coordinate $S_0=0$ and $P_{L_0}=I_{r, n-r}$. For $L\in \eO$ with coordinates $S=S(L)$ we have
$$
P_L-P_{L_0}= . \left[
\begin{array}{cc}
W(S)-1 & W(S)S^*\\
S W(S) & S W(S) S^*
\end{array}
\right]
$$
If we denote by $d_\Lambda$ the metric you defined. Then, setting $W=W(S)$, we get$\DeclareMathOperator{\tr}{tr}$
$$
d_\Lambda(L,L_0)^2=\tr (W-1)\Lambda(W-1) +\tr (SW\Lambda WS^*)
$$
$$
=\tr(\Lambda(W-1)^2)+\tr(\Lambda WS^*SW)
$$
($S^*S$ commutes with $W(S)$)
$$
=\tr(\Lambda(W-1)^2)+\tr(\Lambda W^2S^*S)
$$
$$
=\tr\Lambda \big(\; (W-1)^2+W^2 S^*S\;\big)
$$
For simplicity we set $R=R(S)=S^*S$ Thus
$$
d_\Lambda(L,L_0)^2=\tr\Lambda \big(\; (W-1)^2+W^2 R\;\big).
$$
Note that
$$
W-1=R(1+R)^{-1}=WR,\;\;(W-1)^2=W^2R^2,
$$
$$(W-1)^2+W^2R=W^2(1+R)R=WR.
$$
Hence
$$
d_\Lambda(L,L_0)^2=\tr\Lambda R W=\tr \Lambda R \big(1+R\big)^{-1},\;\;R=R\big(S(L)).
$$
Let us see how this looks, when $r=m$, $n=n+1$. Then $S$ is a $1\times m$ matrix
$$
S=(s_1,\dotsc,s_m)
$$
Then
$$
R(1+R)^{-1}= R- R^2+ O(\Vert R\Vert^3)
$$
Denote by $E_a(\Lambda)$ the ellipsoid
$$
\big\{S\in\bR^m;\;\;\tr \Lambda R(S)\leq a^2\big\}=\big\{S\in\bR^m:\;\;\lambda_1s_1^2+\cdots+\lambda_ms_m^2\leq a^2\big\}.
$$
Then
$$
\lim_{a\searrow 0}\frac{\mu\big( B_a(\Lambda)\big)}{ \mu\big( E_a(\Lambda)\big)}=1,\;\;\forall \Lambda,\;\;\prod_i\lambda_i>0.
\tag{1}
$$
The proof uses the change in coordinates
$$
S=T\Lambda^{-\frac{1}{2}} ,\;\;T=(t_1,\dotsc, t_m),
$$
so
$$
R(S)= \Lambda^{-\frac{1}{2}}R(T)\Lambda^{-\frac{1}{2}}.
$$
If the convergence in (1) were uniform in $\Lambda$, it would confirm your claim in the case $r=n-1$ and supports your initial intuitive comparison with ellipsoids.

Perhaps it would be good to look at the case $m=2$ first. In this case
$$
R(S)=\left[
\begin{array}{cc}s_1^2 & s_1s_2\\
s_1s_2 & s_2^2
\end{array}
\right],\;\; W(S)=\frac{1}{1+s_1^2+s^2} \left[\begin{array}{cc}1+s_2^2 & -s_1s_2\\
-s_1s_2 & 1+s_1^2
\end{array}
\right].
$$
$$
RW= \frac{1}{1+s_1^2+s^2} \left[\begin{array}{cc} s_1^2 &\ast\\
\ast & s_2^2
\end{array}
\right]
$$
$$
\tr(\Lambda RW)=\frac{\lambda_1s_1^2+\lambda_2s_2^2}{1+s_1^2+s^2},\;\;\lambda_1+\lambda_2=1.
$$
As $\lambda_1\searrow 0$ the ball $B_a(\Lambda)$ becomes unbounded in the $S$ coordinates and $\mu$ can no longer be approximated by the Euclidean volume.

The precise form of the invariant volume on $\Gr_r(\bR^m)$ is described in Sec. 9.1.2 of the same notes linked above. The computations in the case $r=2$, $n=3$ are eminently doable and you can test your hypothesis in this case.

**Remark** Suppose that $m=2$ and $a^2=\lambda_1$. Here $a$ is meant to be small. Then
$$
B_a(\Lambda)=\big\{ (a_1,s_2)\in\bR^2;\;\;(1-a^2)s_2^2\leq a^2\big\}
$$
so that in the $(s_1,s_2)$ coordinates $B_a(\Lambda)$ is, up to a $\mu$-negligible set, the strip
$$
|s_2|\leq \frac{a}{\sqrt{1-a^2}}.
$$
For $b^2=1-a^2$ the region $B_b(\Lambda)\cap \eO$, $\lambda_1=a^2$ has the description
$$
a^2s_1^2+(1-a^2)s_2^2)\leq (1-a^2)(1+s_1^2+s_2^2)
$$
i.e.,
$$
(1-a^2)+(1-2a^2)s_1^2\geq 0.
$$
Thus $B_b(\Lambda\cap \eO=\eO$ so
$$
\mu(B_b(\Lambda))= \mu(B_b(\Lambda)\cap \eO)= \mu(\eO)=\mu(\Gr_m(\bR^{m+1}).
$$
If your estimate were to hold uniformly we would have
$$
\mu(B_{\sqrt{\lambda_1}}(\Lambda))=O(\lambda_1).
$$
Given that $B_{\sqrt{\lambda_1}}(\Lambda)$ is the strip $|s_1|\leq \sqrt{\lambda_1}$ its volume is $\geq const.\sqrt{\lambda_1})$.

To see this consider the thin rectangle
$$
D_{\lambda_1}=\{|s_1|\leq 1,\;\;|s_2|\leq \sqrt{\lambda_1}\}.
$$
We write $\mu$ ast
$$
\mu=\rho(s_1,s_2)
$$
and we observe that there exists $C_\mu>0$, independent of $\Lambda$ such that
$$
\rho(s_1,s_2)\geq C_\mu,\;\;\forall |s_1|,|s_2|\leq 1.
$$
Then
$$
\mu(B_{\sqrt{\lambda_1}}(\Lambda))\geq \mu(D_{\lambda_1})\geq C_\mu\sqrt{\lambda_1}.
$$