Is there any connection between QR and SVD of a matrix? Is it possible to draw any parallels between the SVD and QR decomposition of a matrix?
Moreover, for a given matrix $\mathbf{A}\in\mathbb{R}^{n\times m}$, under what conditions, the $\mathbf{U}$ matrix coming from singular value decomposition of $\mathbf{A}$ is equal to $\mathbf{Q}$ matrix obtained via QR-decomposition of $\mathbf{A}$
 A: I don't know if there really is a link between them... Notice that those decompositions are not unique, so the question is a bit unclear. But here is some thoughts on your second question. Maybe it can help you ...
Consider $A \in \Bbb C^{m \times n}$, $m \geq n$. (In the other case, just take the transpose...)
We first fix some notations :
1) A $QR$ decomposition of $A$ is any decomposition $A = QR = Q_1R_1$, where $Q = \begin{pmatrix} Q_1 & Q_2 \end{pmatrix} \in \Bbb C^{m \times m}$ is a unitary matrix, $Q_1 \in \Bbb C^{m \times n}$, and $R = \begin{pmatrix} R_1 \\ 0\end{pmatrix} \in \Bbb C^{m \times n}$, $R_1 \in \Bbb C^{n \times n}$ is upper triangular. We say such a $R$ is upper triangular as well (even if it's not a square matrix).
2) A $SVD$ of $A$ is any decomposition $A = U \Sigma V^*$, where $\Sigma = \begin{pmatrix} \Sigma_1 \\ 0\end{pmatrix} \in \Bbb C^{m \times n}$, $\Sigma_1 \in \Bbb C^{n \times n}$ is diagonal with non negative entries (we say such a $\Sigma$ is diagonal with non negative diagonal entries) and $U \in \Bbb C^{m \times m}$, $V \in \Bbb C^{n \times n}$ are unitary.

Suppose that $A$ has full column rank and let $A = QR$ be a $QR$ decomposition for $A$. There exists a $SVD$ of $A$ such that $Q = U$ if and only if $A^*A$ is diagonal.

Suppose $A^*A$ is diagonal. Then, $A^*A = R^*Q^*QR = R^*R = R_1^*R_1$ so $R_1$ must be diagonal (using the fact that $A$ has full column rank and $R_1$ is upper triangular). Then $A = QR = QRI$ and we can multiply some columns of $R$ and the corresponding rows of $I$ by $-1$ so that we get $A = QR'J = U \Sigma V$, where $R'$ is diagonal with non negative diagonal entries, $Q$ and $J$ are unitary. This is a $SVD$ decomposition for $A$ with $Q = U$.
Now, suppose $Q = U$ for some $SVD$ of $A$. Then we have $\Sigma_1$ is diagonal with strictly positive entries (since $A$ has full column rank) and $\Sigma V^* = R \Rightarrow \Sigma_1 V^* = R_1$ so $V^*$ is upper triangular. Since $V^*$ is unitary and upper triangular, it must be diagonal. But the columns of $V$ form a basis of eigenvectors of $A^*A$. This means $e_i \in \Bbb R^n$ is an eigenvector of $A^*A$ for all $i = 1, ..., n$. Therefore, $A^*A$ is diagonal. 
