[Edit: Allen Hatcher posted an answer while I was writing this one. Both answers seem to use similar ideas. I will leave my answer here anyway.]
$\newcommand{\Diff}{\operatorname{Diff}}$$\newcommand{\ZZ}{\mathbb{Z}}$$\newcommand{\RR}{\mathbb{R}}$$\newcommand{\CC}{\mathbb{C}}$$\newcommand{\connsum}{\mathbin{\\#}}$ $\newcommand{\connsum}{\mathbin{\#}}$$\newcommand{\id}{\mathrm{id}}$$\newcommand{\set}[1]{\lbrace #1 \rbrace}$Reading the answer by Peter Michor gave me an idea for a counter-example, which I explain below. The question still remains as to whether the result is valid for the interior of a compact manifold.
Claim: The path components of a space homotopy equivalent to a CW-complex are all open.
A homotopy equivalence induces a bijection on path components. Consequently, if a space $X$ is homotopy equivalent to a space $Y$, and the path components of $Y$ are open in $Y$, then the path components of $X$ are open in $X$. Finally, observe that any CW-complex is locally path connected, and thus its path components are open.
Construction of the counter-example
It thus suffices to present a smooth manifold $M$ without boundary such that the path components of $\Diff(M)$ are not open. Define $M$ to be the open submanifold of $\RR \times S^1$ given by $$ M = (\RR \times S^1) \setminus (\ZZ \times \set{1}) $$ where $1\in S^1 \subset\CC$. One can also view $M$ as a connected sum of infinitely many punctured spheres $P=S^2\setminus\set{(1,0,0)}$: $$ M \cong \, \cdots \connsum P \connsum P \connsum P \connsum \cdots $$
Proof that the path component of $\id_M$ is not open in $\Diff(M)$
Pick a neighbourhood $U$ of $\id_M$ in $\Diff(M)$. I will describe a diffeomorphism $\varphi\in U$ which is not in the path component of $\id_M$ in $\Diff(M)$, thus concluding the proof. By definition of the compact-open topology on $\Diff(M)$, there exists a compact subspace $K$ of $M$ such that a given diffeomorphism of $M$ is in $U$ if it is the identity on $K$. Let $n$ be a positive integer large enough so that $K\subset [-n,n]\times S^1$. Now we define the required diffeomorphism $\varphi$ of $M$: $$ \varphi(t,x) = \bigl( t , e^{i\cdot\theta(t-n)} x \bigr) $$ where $\theta:\RR\to\RR$ is any smooth function which is identically zero on $(-\infty,0]$, and equals $2\pi$ on $[1,+\infty)$. Viewing $M$ as the connected sum of infinitely many punctured spheres, the diffeomorphism $\varphi$ is the result of applying a Dehn twist at the junction cylinder joining the spheres numbered $n$ and $n+1$.
It is easy to see that $\varphi:M\to M$ is not even homotopic to the identity: in fact, the homomorphism induced by $\varphi$ on $\pi_1(M)$ (based at the point $(n,-1)$) is not conjugate to the identity. Here is a simple way to see this:
The group $\pi_1 M$ is free on infinitely many generators.
The homomorphism $\pi_1 \varphi$ coincides with the identity on all loops in $M$ contained in $(-\infty,n]\times S^1$. In particular, $\pi_1 \varphi$ fixes two distinct (actually, infinitely many) free generators of $\pi_1 M$.
The homomorphism $\pi_1 \varphi$ is not the identity homomorphism on $\pi_1 M$: a loop going once around the puncture $(n+1,1)\in\RR\times S^1$ is not fixed by $\pi_1 \varphi$.
In a free group, conjugation by a non-identity element can fix at most one of the free generators.