Parametric representation of orthogonal matrices

Every orthogonal matrix with determinant $1$ has the form $\exp(X)$ with $X$ skew-symmetric. Another representation of orthogonal matrices is the Cayley parameterisation: $(I+X)(I-X)^{-1}$ is orthogonal whenever $X$ is skew-symmetric. This produces all orthogonal matrices of determinant $1$ which do not have $-1$ as an eigenvalue.

As you said, if $X$ is a skew-symmetric matrix then $e^X$ is orthogonal, and precisely $e^X$ is a special orthogonal matrix, i.e. with determinant equal to $+1$: $$\det(e^X)=e^{\operatorname{Tr}(X)}=e^0=1$$ Conversely, for $O\in SO_n(\Bbb R)$, there is an orthonormal basis in which $O$ is similar to $$S=\operatorname{diag}(I_p,R(\theta_1),\ldots,R(\theta_s))$$ where $$R(\theta_i)=\begin{pmatrix}\cos\theta_i&-\sin\theta_i\\\sin\theta_i&\cos\theta_i\end{pmatrix}$$ Moreover, we have $\exp(J_i)=R(\theta_i)$ with $$J_i=\begin{pmatrix}0&-\theta_i\\\theta_i&0\end{pmatrix}$$ so we define the skew-symmetric matrix $$A=\operatorname{diag}(0_p, J_1,\ldots,J_s)$$ and we get $e^A=S$.

As Lord Shark of the Unknown already told, the answer is negative for $O(n,\mathbb{R})$, but it is affirmative for $SO(n,\mathbb{R})$. That the answer is negative in the case of $O(n,\mathbb{R})$ follows from the fact that it is not connected, whereas the space of all skew-symmetric matrices is (together with the fact that the exponential map is continuous).

Actually, if $K$ is a compact and connected matrix group and if$$\mathfrak{k}=\left\{X\in M(n,\mathbb{R})\,|\,(\forall t\in\mathbb{R}):\exp(tX)\in K\right\},$$then $\exp(\mathfrak{k})=K$.