$\newcommand{\Q}{\mathbb Q}\newcommand{\erf}{\operatorname{erf}}$(This answer had been posted before I saw Command Master's answer. I am leaving it here, since it contains more and/or different details; in particular, here the argument that the function $\erf$ is non-elementary is not used.)

The answer is yes. Moreover, the value of the integral
\begin{equation*}
I:=\int_0^1 e^{-t^2}\,dt
\end{equation*}
is transcendental.

To show this, let us use some of the comments, with added details:

**Step 0:** For real $x>0$, let
\begin{equation*}
y(x):=\frac{\sqrt\pi}{2\sqrt x}\,\erf\sqrt x=\sum_{n=0}^\infty c_n\frac{x^n}{n!},
\end{equation*}
with
\begin{equation*}
c_n:=\frac{(-1)^n}{2n+1}
\end{equation*}
and $y(0):=1$, so that
\begin{equation*}
I=y(1).
\end{equation*}

**Step 1:** The function $y$ is an E-function over $\Q$.

Indeed, of the three bulleted conditions listed in the definition of an E-function in the Wikipedia article, only the last, third one is nontrivial. To verify it, let $q_n$ be the least common multiple of $1,\dots,2n+1$. Then, in view of the prime number theorem,

\begin{multline*}
q_n=\prod_{p\le2n+1}p^{\lfloor\log_p(2n+1)\rfloor}
\le\prod_{p\le2n+1}p^{\log_p(2n+1)} \\
=\prod_{p\le2n+1}(2n+1)=(2n+1)^{\pi_{2n+1}} \\ =(2n+1)^{(1+o(1))(2n+1)/\ln(2n+1)} \\
=e^{(1+o(1))(2n+1)}=O(n^{n\varepsilon})
\end{multline*}
for each real $\varepsilon>0$, where the product $\prod_{p\le2n+1}$ is over all primes $p\le2n+1$ and $\pi_k$ is the number of primes $\le k$.

**Step 2:** Letting $y_1(x):=y(x)$ and $y_2(x):=y'(x)$, and using Command Master's comment, we get the system of ODEs
\begin{equation*}
y'_1(x)=y_2(x),\quad y'_2(x)=-\frac1{2x}\,y_1(x)-\Big(1+\frac3{2x}\Big)y_2(x). \tag{1}\label{1}
\end{equation*}
We now want to use the Siegel–Shidlovsky theorem. The only possibly nontrivial condition of this theorem to check here is that the "functions" $y_1(x)$ and $y_2(x)$ are algebraically independent over $\Q(x)$ or, equivalently, that the "functions"
\begin{equation*}
v_1(t):=y_1(t^2)=\frac{\sqrt\pi}{2t}\,\erf t\quad\text{and}\quad v_2(t):=y_2(t^2)=\sqrt\pi\,\frac{t\erf' t-\erf t}{2t^3}
\end{equation*}
are algebraically independent over $\Q(t^2)$. So, it is enough to show that $\erf t$ and $\erf' t$ are algebraically independent over $\Q(t)$.

To do this, suppose that
\begin{equation*}
\sum_{(j,k)\in F}P_{j,k}\,\erf^{\,j} (\erf')^k=0
\end{equation*}
for some nonempty finite subset $F$ of the set $\{0,1,\dots\}^2$ and some polynomial functions $P_{j,k}$. The latter identity can be rewritten as
\begin{equation*}
\sum_{j=0}^m Q_j \erf^{\,j}=0 \tag{2}\label{2}
\end{equation*}
for some integer $m\ge0$, where each $Q_j(t)$ is rational in $(t,\erf'(t))$ and $Q_m=1$. Differentiating \eqref{2}, we get an identity of the form \eqref{2}, but with $m-1$ in place of $m$ (if $m\ge1$). So, without loss of generality $m=0$. So, $Q_0=0$. Similarly repeatedly rewriting and differentiating, we conclude that all the coefficients of the $(\erf')^k$'s in $Q_0$ are $0$. This confirms that $\erf t$ and $\erf' t$ are algebraically independent over $\Q(t)$, and hence
$y_1(x)$ and $y_2(x)$ are algebraically independent over $\Q(x)$.

**Step 3:** By the Siegel–Shidlovsky theorem, $I=y(1)$ and $\frac1e=y'(1)+\frac12\,y(1)$ are algebraically independent. In particular, it follows that $I$ and $e$ are transcendental and hence irrational. $\quad\Box$

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