The answer to your question is positive and follows immediately from, e.g., Corollary 4 of

C. A. León and F. Perron (2003), Extremal properties of sums of Bernoulli random variables, Statistics and Probability Letters, vol. 62, 345–354.

A slight specialization of the corollary states:

Let $\newcommand{\E}{\mathbb E} Y = Y_1 + \cdots + Y_n$ be a sum of iid random variables taking values in $[0,1]$ with mean $\E Y_i = \mu$ and let $X \sim \mathrm{Bin}(n,\mu)$. For any convex function $g : [0,n] \to \mathbb R$, $$ \E g(Y) \leq \E g(X) \ . $$

Your result follows by noting that $g_a(x) = -\min(x,a)$ is convex for $a > 0$.

The proof is easy and direct, so we may as well give a version of it here as several short lemmas. In what follows below, we assume $(Y_i)$ are iid on $[0,1]$ with distribution function $F$ and mean $\mu$ and that $(X_i)$ are iid Bernoulli random variables also with mean $\mu$.

Lemma 1: If $Y \sim F$ and $X \sim \mathrm{Ber}(\mu)$, then for any convex $g : [0,1] \to \mathbb R$, $\E g(Y) \leq \E g(X)$.

Proof: $\E g(Y) \leq g(1) \E Y + g(0) \E (1-Y) = \E g(X)$.

Lemma 2: For any convex $g$, $\E g(Y_1 + Y_2) \leq \E g(X_1 + X_2)$.

Proof: Assume wlog that $(Y_1,Y_2)$ is independent of $(X_1,X_2)$. If $g$ is convex, then so is $g(y+\cdot)$, hence $$\E g(Y_1 + Y_2) = \int_0^1 \E g(y + Y_2) \ \mathrm dF \leq \int_0^1 \E g(y + X_2) \ \mathrm dF = \E g(Y_1 + X_2) \leq \E g(X_1 + X_2) \ . $$

Corollary: Let $Y = Y_1 + \cdots + Y_n$ and $X = X_1 + \cdots X_n$, where $X \sim \mathrm{Bin}(n,\mu)$ as in the problem. Then, then for any convex $g$, $\E g(Y) \leq \E g(X)$.

Proof: Extend the previous lemma by induction.

The desired result now follows since $g(x,a) = \min(x,a)$ is concave in $x$ for $a > 0$.

The answer to your question is positive and, for example, follows immediately from Corollary 4 of

C. A. León and F. Perron (2003), Extremal properties of sums of Bernoulli random variables, Statistics and Probability Letters, vol. 62, 345–354.

A slight specialization of the corollary states:

Let $\newcommand{\E}{\mathbb E} Y = Y_1 + \cdots + Y_n$ be a sum of iid random variables taking values in $[0,1]$ with mean $\E Y_i = \mu$ and let $X \sim \mathrm{Bin}(n,\mu)$. For any convex function $g : [0,n] \to \mathbb R$, $$ \E g(Y) \leq \E g(X) \ . $$

Your result follows by noting that $g(x,a) = -\min(x,a)$ is convex in $x$ for any $a \in \mathbb R$.

The paper appeals to other references, but an easy and direct proof can be constructed, so we may as well give a version of it here as several short lemmas. 

In what follows below, we assume $(Y_i)$ are iid on $[0,1]$ with distribution function $F$ and mean $\mu$ and that $(X_i)$ are iid Bernoulli random variables also with mean $\mu$. The function $g$ is assumed to be an arbitrary convex function defined on the appropriate domain.

Lemma 1: If $Y \sim F$ and $X \sim \mathrm{Ber}(\mu)$, then $\E g(Y) \leq \E g(X)$.

Proof: $Y = 1 \cdot Y + 0 \cdot (1-Y)$, so, by convexity, $\E g(Y) \leq g(1) \E Y + g(0) \E (1-Y) = \E g(X)$.

Lemma 2: $\E g(Y_1 + Y_2) \leq \E g(X_1 + X_2)$.

Proof: Assume wlog that $(Y_1,Y_2)$ is independent of $(X_1,X_2)$. If $g$ is convex, then so is $g(y+\cdot)$, hence $$\E g(Y_1 + Y_2) = \int_0^1 \E g(y + Y_2) \ \mathrm dF \leq \int_0^1 \E g(y + X_2) \ \mathrm dF = \E g(Y_1 + X_2) \leq \E g(X_1 + X_2) \ . $$

Corollary: Let $Y = Y_1 + \cdots + Y_n$ and $X = X_1 + \cdots X_n$, where $X \sim \mathrm{Bin}(n,\mu)$ as in the problem. Then, $\E g(Y) \leq \E g(X)$.

Proof: Extend the previous lemma by induction.

The desired result now follows by taking the aforementioned choice for $g$ with $a = n \mu$.