Undeleting and editing my own answer, not to bump the question up again, but to try to close / settle it as I think there are some interesting subtleties in it.
I'll also briefly address @MainCom 's counterexample to show that in fact it isn't a counterexample.
As I'd like to use the mean value theorem I'll assume that the asset process is continuous on $[0,T]$. For example a local stochastic volatility model without asset price jumps will satisfy this condition.
For simplicity I've set the risk-free rate and dividend yield to zero.
We will be interested in vanilla options
\begin{equation}
C\left(S_t,t,K,T\right) := E_t \left[ \left(S_T - K\right)_+ \right],
\end{equation}
and Asian options
\begin{equation}
C\left(A_t,t,K,T\right) := E_t \left[ \left(A_T - K\right)_+ \right],
\end{equation}
with
\begin{equation}
A_t := E_t \left[ \frac1T \int_0^T S_u \, du \right] .
\end{equation}
Let $BS\left(S_t,t,K,T,I_S (K)\right)$ denote the Black-Scholes (BS) price of a vanilla option with implied volatility (IV) $I_S (K)$, and $BS\left(A_t,t,K,T,I_A (K)\right)$ the BS price of an Asian option with IV $I_A (K)$. These IVs are defined by
\begin{align}
BS\left(S_t,t,K,T,I_S(K)\right) &:= C\left(S_t,t,K,T\right), \\
BS\left(A_t,t,K,T,I_A(K)\right) &:= C \left(A_t,t,K,T\right).
\end{align}
Lastly, recall also the mean-value theorem for integrals: Let $f:[a,b]\rightarrow \mathbb{R}$ be a continuous function. Then there exists at least one $x\in[a,b]]$ such that
$$
f(x) = \frac{1}{b-a} \int_a^b f(u)\, du.
$$
Proposition:
An upper bound on the price $BS\left(A_t,t,K,T,I_A (K)\right)$ of an Asian option is
\begin{equation}
BS\left(A_t,t,K,T,I_A(K)\right) \leq \lambda \, BS\left(S_t,t,\lambda^{-1}K' ,T,I_S(\lambda^{-1}K')\right),
\end{equation}
with $\lambda = \frac{T-t}{T}$ and $K' = K - \frac1T \int_0^t S_u \, du$.
Proof:
We can write
\begin{align*}
BS\left(A_t,t,K,T,I_A (K)\right) &:= E_t \left[ \left(A_T - K\right)_+ \right] \\
&= E_t \left[ \left(\frac1T \int_0^T S_u \, du - K \right)_+ \right] \\
&= E_t \left[ \left(\frac{\lambda}{T-t} \int_t^T S_u \, du - K' \right)_+ \right]
\end{align*}
with $\lambda = \frac{T-t}{T}$ and $K' = K - \frac1T \int_0^t S_u \, du$. According to the mean-value theorem, for each path of the asset, there exists at least one $\tau\in [t,T]$ such that
$$
S_\tau = \frac{1}{T-t} \int_t^T S_u du.
$$
Let $\tau^*$ be the first such $\tau$. It is clear that each $\tau \in [t,T]$ is a random variable, and in particular $\tau^* \in [t,T]$ is a random variable. The problem of determining the price of an Asian option can then be re-cast in the following form:
$$
BS\left(A_t,t,K,T,I_A (K)\right) = \lambda E_t \left[ \left(S_{\tau^*} - \lambda^{-1}K' \right)_+ \right].
$$
Denote by $q(r)$ the distribution of $\tau^*$. Then
\begin{align*}
BS\left(A_t,t,K,T,I_A (K)\right) &= \lambda \int_t^T E_t\left[ \left(S_{\tau^*} - \lambda^{-1} K' \right)_+ | \tau^* = r \right] q(r)\, dr
\end{align*}
Now,
\begin{align*}
E_t\left[ \left(S_{\tau^*} - \lambda^{-1} K' \right)_+ | \tau^* = r \right] &= E_t\left[ \left(E_{\tau^*}(S_T) - \lambda^{-1} K' \right)_+ | \tau^* = r \right] \\
&\leq E_t\left[E_{\tau^*} \left(S_T - \lambda^{-1} K' \right)_+ | \tau^* = r \right] \\
&= E_t\left[\left(S_T - \lambda^{-1} K' \right)_+ | \tau^* = r \right] \\
&=E_t\left[\left(S_T - \lambda^{-1} K' \right)_+\right]
\end{align*}
where the inequality follows from Jensen's inequality, and clearly $S_T$ is independent of $\tau^*$.
Hence,
\begin{align*}
BS\left(A_t,t,K,T,I_A (K)\right) &\leq \lambda \int_t^T E_t\left[ \left(S_T - \lambda^{-1} K' \right)_+ \right] q(r)\, dr\\
&= \lambda \, E_t\left[ \left(S_T - \lambda^{-1} K' \right)_+ \right] \\
&= \lambda \, BS\left(S_t,t,\lambda^{-1}K' ,T,I_S (\lambda^{-1}K')\right).
\end{align*}
Corollary:
The IV of a freshly minted Asian option is bounded above by the IV of a vanilla option with the same strike and time to maturity.
Proof:
For a freshly minted Asian option $t=0$ and thus $\lambda = 1$ and $K=K'$.
As for MainCom's counterexample: Even though it is true that calendar arbitrage cannot in general be applied to random times, it is not a counterexample since the maximum of an asset on $[0,T]$ can't be written as an integral with integral limits equal to $0$ and $T$. This means that the mean value theorem can't even be applied to MainCom's example rendering the whole random time argument non-applicable.
Note that the bounds are in line with the more straightforward derivation given for instance here.
However, I thought applying mean value theorem in this context is interesting as well.
Afterthought: The whole derivation can be shortened to
\begin{align*}
BS\left(A_t,t,K,T,I_A (K)\right) &= \lambda E_t \left[ \left(S_{\tau^*} - \lambda^{-1}K' \right)_+ \right] \\
& = \lambda E_t \left[ \left(E_{\tau^*}(S_T) - \lambda^{-1}K' \right)_+ \right] \\
&\leq \lambda E_t \left[ E_{\tau^*} \left(S_T - \lambda^{-1}K' \right)_+ \right] \\
&= \lambda E_t \left[ \left(S_T - \lambda^{-1}K' \right)_+ \right].
\end{align*}
